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Solve-for-n-i-n-1-n-C-i-2-i-65-n-Z-where-zero-is-included-




Question Number 57688 by Tawa1 last updated on 10/Apr/19
  Solve for  n:        Σ_i ^(n − 1)     ^n C_i  2^i   =  65,            n ∈ Z^+ .    where  zero is     included
$$\:\:\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{n}:\:\:\:\:\:\:\:\:\underset{\mathrm{i}} {\overset{\mathrm{n}\:−\:\mathrm{1}} {\sum}}\:\:\:\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{i}} \:\mathrm{2}^{\mathrm{i}} \:\:=\:\:\mathrm{65},\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{n}\:\in\:\mathbb{Z}^{+} .\:\:\:\:\mathrm{where}\:\:\mathrm{zero}\:\mathrm{is}\: \\ $$$$\:\:\mathrm{included} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Apr/19
question  not clear ..  Σ_(i=0) ^(n−1) C_i ^n 2^i ←is it the question
$${question}\:\:{not}\:{clear}\:.. \\ $$$$\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{C}_{{i}} ^{{n}} \mathrm{2}^{{i}} \leftarrow{is}\:{it}\:{the}\:{question} \\ $$
Commented by Tawa1 last updated on 10/Apr/19
Yes sir.
$$\mathrm{Yes}\:\mathrm{sir}.\: \\ $$
Commented by Abdo msup. last updated on 10/Apr/19
first we have Σ_(i=0) ^(n−1)  C_n ^i  2^i  =Σ_(i=0) ^n  C_n ^i  2^i  −C_n ^n  2^n   =3^n  −2^n  (we suppose that the sum begins from 0)  ⇒(e) ⇒3^n  −2^n  =65  ⇒ n=4 .
$${first}\:{we}\:{have}\:\sum_{{i}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{C}_{{n}} ^{{i}} \:\mathrm{2}^{{i}} \:=\sum_{{i}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{i}} \:\mathrm{2}^{{i}} \:−{C}_{{n}} ^{{n}} \:\mathrm{2}^{{n}} \\ $$$$=\mathrm{3}^{{n}} \:−\mathrm{2}^{{n}} \:\left({we}\:{suppose}\:{that}\:{the}\:{sum}\:{begins}\:{from}\:\mathrm{0}\right) \\ $$$$\Rightarrow\left({e}\right)\:\Rightarrow\mathrm{3}^{{n}} \:−\mathrm{2}^{{n}} \:=\mathrm{65}\:\:\Rightarrow\:{n}=\mathrm{4}\:. \\ $$$$ \\ $$
Commented by Tawa1 last updated on 10/Apr/19
God bless you sirs,  is there anyway to find n from  3^n  − 2^n   =  65
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sirs},\:\:\mathrm{is}\:\mathrm{there}\:\mathrm{anyway}\:\mathrm{to}\:\mathrm{find}\:\mathrm{n}\:\mathrm{from}\:\:\mathrm{3}^{\mathrm{n}} \:−\:\mathrm{2}^{\mathrm{n}} \:\:=\:\:\mathrm{65} \\ $$
Commented by Tawa1 last updated on 10/Apr/19
Sir, please i need the rules of summation,  expecially the sum  that has  something like    Σ_(i = 0) ^(n − 1)    how to split them
$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{i}\:\mathrm{need}\:\mathrm{the}\:\mathrm{rules}\:\mathrm{of}\:\mathrm{summation},\:\:\mathrm{expecially}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\mathrm{that}\:\mathrm{has}\:\:\mathrm{something}\:\mathrm{like}\:\:\:\:\underset{\mathrm{i}\:=\:\mathrm{0}} {\overset{\mathrm{n}\:−\:\mathrm{1}} {\sum}}\:\:\:\mathrm{how}\:\mathrm{to}\:\mathrm{split}\:\mathrm{them} \\ $$
Commented by Tawa1 last updated on 10/Apr/19
Because have been trying to understand how to get     3^n  − 2^n   =  65.  I did not understand yet.
$$\mathrm{Because}\:\mathrm{have}\:\mathrm{been}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{how}\:\mathrm{to}\:\mathrm{get} \\ $$$$\:\:\:\mathrm{3}^{\mathrm{n}} \:−\:\mathrm{2}^{\mathrm{n}} \:\:=\:\:\mathrm{65}.\:\:\mathrm{I}\:\mathrm{did}\:\mathrm{not}\:\mathrm{understand}\:\mathrm{yet}. \\ $$
Commented by Tawa1 last updated on 10/Apr/19
Sorry for disturbing sir
$$\mathrm{Sorry}\:\mathrm{for}\:\mathrm{disturbing}\:\mathrm{sir} \\ $$
Commented by Tawa1 last updated on 10/Apr/19
or textbook i can learn it.
$$\mathrm{or}\:\mathrm{textbook}\:\mathrm{i}\:\mathrm{can}\:\mathrm{learn}\:\mathrm{it}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Apr/19
S=C_0 ^n 2^0 +C_1 ^n 2^1 +C_2 ^n 2^2 +....+C_(n−1) ^n 2^(n−1)   now look here  (x+2)^n =C_0 ^n x^(n−0) 2^0 +C_1 ^n x^(n−1) 2^1 +C_2 ^n x^(n−2) 2^2 +...+C_(n−1) ^n x^1 2^(n−1) +C_n ^n x^(n−n) 2^n   now put x=1 both side  3^n =[C_0 ^n 2^0 +C_1 ^n 2^1 +...+C_(n−1) ^n 2^(n−1) ]+C_n ^n 2^n ←look here  3^n =S+C_n ^n 2^n   now C_n ^n =((n!)/(n!(n−n)!))=1  so  3^n =S+2^n ×1  S=3^n −2^n   so Σ_(i=0) ^(n−1) C_i ^n_  2^i =S=3^n −2^n   3^n −2^n =65  by lnspection  3^n −2^n =81−16  3^n −2^n =3^4 −2^4  →n=4
$${S}={C}_{\mathrm{0}} ^{{n}} \mathrm{2}^{\mathrm{0}} +{C}_{\mathrm{1}} ^{{n}} \mathrm{2}^{\mathrm{1}} +{C}_{\mathrm{2}} ^{{n}} \mathrm{2}^{\mathrm{2}} +….+{C}_{{n}−\mathrm{1}} ^{{n}} \mathrm{2}^{{n}−\mathrm{1}} \\ $$$${now}\:{look}\:{here} \\ $$$$\left({x}+\mathrm{2}\right)^{{n}} ={C}_{\mathrm{0}} ^{{n}} {x}^{{n}−\mathrm{0}} \mathrm{2}^{\mathrm{0}} +{C}_{\mathrm{1}} ^{{n}} {x}^{{n}−\mathrm{1}} \mathrm{2}^{\mathrm{1}} +{C}_{\mathrm{2}} ^{{n}} {x}^{{n}−\mathrm{2}} \mathrm{2}^{\mathrm{2}} +…+{C}_{{n}−\mathrm{1}} ^{{n}} {x}^{\mathrm{1}} \mathrm{2}^{{n}−\mathrm{1}} +{C}_{{n}} ^{{n}} {x}^{{n}−{n}} \mathrm{2}^{{n}} \\ $$$${now}\:{put}\:{x}=\mathrm{1}\:{both}\:{side} \\ $$$$\mathrm{3}^{{n}} =\left[{C}_{\mathrm{0}} ^{{n}} \mathrm{2}^{\mathrm{0}} +{C}_{\mathrm{1}} ^{{n}} \mathrm{2}^{\mathrm{1}} +…+{C}_{{n}−\mathrm{1}} ^{{n}} \mathrm{2}^{{n}−\mathrm{1}} \right]+{C}_{{n}} ^{{n}} \mathrm{2}^{{n}} \leftarrow{look}\:{here} \\ $$$$\mathrm{3}^{{n}} ={S}+{C}_{{n}} ^{{n}} \mathrm{2}^{{n}} \\ $$$${now}\:{C}_{{n}} ^{{n}} =\frac{{n}!}{{n}!\left({n}−{n}\right)!}=\mathrm{1} \\ $$$${so} \\ $$$$\mathrm{3}^{{n}} ={S}+\mathrm{2}^{{n}} ×\mathrm{1} \\ $$$${S}=\mathrm{3}^{{n}} −\mathrm{2}^{{n}} \\ $$$${so}\:\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{C}_{{i}} ^{{n}_{} } \mathrm{2}^{{i}} =\boldsymbol{{S}}=\mathrm{3}^{{n}} −\mathrm{2}^{{n}} \\ $$$$\mathrm{3}^{{n}} −\mathrm{2}^{{n}} =\mathrm{65} \\ $$$${by}\:{lnspection} \\ $$$$\mathrm{3}^{{n}} −\mathrm{2}^{{n}} =\mathrm{81}−\mathrm{16} \\ $$$$\mathrm{3}^{{n}} −\mathrm{2}^{{n}} =\mathrm{3}^{\mathrm{4}} −\mathrm{2}^{\mathrm{4}} \:\rightarrow{n}=\mathrm{4} \\ $$$$ \\ $$
Commented by Tawa1 last updated on 10/Apr/19
Ohh, I appreciate your time and effort sir.  God bless you sir
$$\mathrm{Ohh},\:\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{and}\:\mathrm{effort}\:\mathrm{sir}.\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by Tawa1 last updated on 10/Apr/19
I really appreciate sir
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir} \\ $$
Commented by Tawa1 last updated on 10/Apr/19
And i understand very well
$$\mathrm{And}\:\mathrm{i}\:\mathrm{understand}\:\mathrm{very}\:\mathrm{well} \\ $$
Answered by mr W last updated on 10/Apr/19
assumed you mean  Σ_(i=0) ^(n − 1)     ^n C_i  2^i   =  65   LHS=(1+2)^n −2^n   (1+2)^n −2^n =65  3^n −2^n =65  ⇒n=4
$${assumed}\:{you}\:{mean} \\ $$$$\underset{\mathrm{i}=\mathrm{0}} {\overset{\mathrm{n}\:−\:\mathrm{1}} {\sum}}\:\:\:\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{i}} \:\mathrm{2}^{\mathrm{i}} \:\:=\:\:\mathrm{65}\: \\ $$$${LHS}=\left(\mathrm{1}+\mathrm{2}\right)^{{n}} −\mathrm{2}^{{n}} \\ $$$$\left(\mathrm{1}+\mathrm{2}\right)^{{n}} −\mathrm{2}^{{n}} =\mathrm{65} \\ $$$$\mathrm{3}^{{n}} −\mathrm{2}^{{n}} =\mathrm{65} \\ $$$$\Rightarrow{n}=\mathrm{4} \\ $$
Commented by Tawa1 last updated on 10/Apr/19
How is left hand side =  (1 + 2)^n  − 2^n    sir  ??
$$\mathrm{How}\:\mathrm{is}\:\mathrm{left}\:\mathrm{hand}\:\mathrm{side}\:=\:\:\left(\mathrm{1}\:+\:\mathrm{2}\right)^{\mathrm{n}} \:−\:\mathrm{2}^{\mathrm{n}} \:\:\:\mathrm{sir}\:\:?? \\ $$
Commented by mr W last updated on 10/Apr/19
(1+2)^n =Σ_(i=0) ^n C_i ^n 2^i =Σ_(i=0) ^(n−1) C_i ^n 2^i +2^n =LHS+2^n
$$\left(\mathrm{1}+\mathrm{2}\right)^{{n}} =\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{i}} ^{{n}} \mathrm{2}^{{i}} =\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{C}_{{i}} ^{{n}} \mathrm{2}^{{i}} +\mathrm{2}^{{n}} ={LHS}+\mathrm{2}^{{n}} \\ $$
Commented by Tawa1 last updated on 10/Apr/19
God bless you sir. any step to get n = 4 ?
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{any}\:\mathrm{step}\:\mathrm{to}\:\mathrm{get}\:\mathrm{n}\:=\:\mathrm{4}\:? \\ $$
Commented by mr W last updated on 10/Apr/19
try and see...
$${try}\:{and}\:{see}… \\ $$
Commented by Tawa1 last updated on 10/Apr/19
Trier and error ?
$$\mathrm{Trier}\:\mathrm{and}\:\mathrm{error}\:? \\ $$
Commented by Tawa1 last updated on 10/Apr/19
I found a method i used to get  n = 4,  from      3^n  − 2^n   =  65.  I used algebra  representation.  But am still finding it difficult to understand.  how to get    3^n  − 2^n   =  65     from the relation.
$$\mathrm{I}\:\mathrm{found}\:\mathrm{a}\:\mathrm{method}\:\mathrm{i}\:\mathrm{used}\:\mathrm{to}\:\mathrm{get}\:\:\mathrm{n}\:=\:\mathrm{4},\:\:\mathrm{from}\:\:\:\:\:\:\mathrm{3}^{\mathrm{n}} \:−\:\mathrm{2}^{\mathrm{n}} \:\:=\:\:\mathrm{65}.\:\:\mathrm{I}\:\mathrm{used}\:\mathrm{algebra} \\ $$$$\mathrm{representation}. \\ $$$$\mathrm{But}\:\mathrm{am}\:\mathrm{still}\:\mathrm{finding}\:\mathrm{it}\:\mathrm{difficult}\:\mathrm{to}\:\mathrm{understand}.\:\:\mathrm{how}\:\mathrm{to}\:\mathrm{get} \\ $$$$\:\:\mathrm{3}^{\mathrm{n}} \:−\:\mathrm{2}^{\mathrm{n}} \:\:=\:\:\mathrm{65}\:\:\:\:\:\mathrm{from}\:\mathrm{the}\:\mathrm{relation}.\: \\ $$
Commented by Tawa1 last updated on 10/Apr/19
Maybe i need sigma notation rules. How to apply
$$\mathrm{Maybe}\:\mathrm{i}\:\mathrm{need}\:\mathrm{sigma}\:\mathrm{notation}\:\mathrm{rules}.\:\mathrm{How}\:\mathrm{to}\:\mathrm{apply} \\ $$
Commented by Tawa1 last updated on 10/Apr/19
 3^n  − 2^n   =  65      [3^((n/2)) ]^2  − [2^((n/2)) ]^2   =  65  3^(n/2)   =  m   and     2^(n/2)  = y        m^2  − y^2   =  65         (m + y)(m − y)  =  13 × 5          m + y  = 13          m − y  =  5  m  =  9,  y = 4     3^(n/2)   =  3^2      and      2^(n/2)   =  2^2         n = 4   twice.
$$\:\mathrm{3}^{\mathrm{n}} \:−\:\mathrm{2}^{\mathrm{n}} \:\:=\:\:\mathrm{65} \\ $$$$\:\:\:\:\left[\mathrm{3}^{\left(\mathrm{n}/\mathrm{2}\right)} \right]^{\mathrm{2}} \:−\:\left[\mathrm{2}^{\left(\mathrm{n}/\mathrm{2}\right)} \right]^{\mathrm{2}} \:\:=\:\:\mathrm{65} \\ $$$$\mathrm{3}^{\mathrm{n}/\mathrm{2}} \:\:=\:\:\mathrm{m}\:\:\:\mathrm{and}\:\:\:\:\:\mathrm{2}^{\mathrm{n}/\mathrm{2}} \:=\:\mathrm{y} \\ $$$$\:\:\:\:\:\:\mathrm{m}^{\mathrm{2}} \:−\:\mathrm{y}^{\mathrm{2}} \:\:=\:\:\mathrm{65} \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{m}\:+\:\mathrm{y}\right)\left(\mathrm{m}\:−\:\mathrm{y}\right)\:\:=\:\:\mathrm{13}\:×\:\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{m}\:+\:\mathrm{y}\:\:=\:\mathrm{13} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{m}\:−\:\mathrm{y}\:\:=\:\:\mathrm{5} \\ $$$$\mathrm{m}\:\:=\:\:\mathrm{9},\:\:\mathrm{y}\:=\:\mathrm{4} \\ $$$$\:\:\:\mathrm{3}^{\mathrm{n}/\mathrm{2}} \:\:=\:\:\mathrm{3}^{\mathrm{2}} \:\:\:\:\:\mathrm{and}\:\:\:\:\:\:\mathrm{2}^{\mathrm{n}/\mathrm{2}} \:\:=\:\:\mathrm{2}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{n}\:=\:\mathrm{4}\:\:\:\mathrm{twice}. \\ $$$$ \\ $$
Commented by Tawa1 last updated on 10/Apr/19
But am still comfused on how to get   3^n  − 2^n   from the sum
$$\mathrm{But}\:\mathrm{am}\:\mathrm{still}\:\mathrm{comfused}\:\mathrm{on}\:\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\:\:\mathrm{3}^{\mathrm{n}} \:−\:\mathrm{2}^{\mathrm{n}} \:\:\mathrm{from}\:\mathrm{the}\:\mathrm{sum} \\ $$
Commented by MJS last updated on 10/Apr/19
this isn′t a method, it works here just by chance  3^n −2^n =91  (3^(n/2) )^2 −(2^(n/2) )^2 =91  (3^(n/2) −2^(n/2) )(3^(n/2) +2^(n/2) )=7×13  3^(n/2) −2^(n/2) =7  3^(n/2) +2^(n/2) =13  ⇒ 3^(n/2) =10 ∧ 2^(n/2) =3  ⇒ n=((2ln 10)/(ln 3)) ∧ n=((2ln 3)/(ln 2))  while in this case n≈4.28235
$$\mathrm{this}\:\mathrm{isn}'\mathrm{t}\:\mathrm{a}\:\mathrm{method},\:\mathrm{it}\:\mathrm{works}\:\mathrm{here}\:\mathrm{just}\:\mathrm{by}\:\mathrm{chance} \\ $$$$\mathrm{3}^{{n}} −\mathrm{2}^{{n}} =\mathrm{91} \\ $$$$\left(\mathrm{3}^{{n}/\mathrm{2}} \right)^{\mathrm{2}} −\left(\mathrm{2}^{{n}/\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{91} \\ $$$$\left(\mathrm{3}^{{n}/\mathrm{2}} −\mathrm{2}^{{n}/\mathrm{2}} \right)\left(\mathrm{3}^{{n}/\mathrm{2}} +\mathrm{2}^{{n}/\mathrm{2}} \right)=\mathrm{7}×\mathrm{13} \\ $$$$\mathrm{3}^{{n}/\mathrm{2}} −\mathrm{2}^{{n}/\mathrm{2}} =\mathrm{7} \\ $$$$\mathrm{3}^{{n}/\mathrm{2}} +\mathrm{2}^{{n}/\mathrm{2}} =\mathrm{13} \\ $$$$\Rightarrow\:\mathrm{3}^{{n}/\mathrm{2}} =\mathrm{10}\:\wedge\:\mathrm{2}^{{n}/\mathrm{2}} =\mathrm{3} \\ $$$$\Rightarrow\:{n}=\frac{\mathrm{2ln}\:\mathrm{10}}{\mathrm{ln}\:\mathrm{3}}\:\wedge\:{n}=\frac{\mathrm{2ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{2}} \\ $$$$\mathrm{while}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:{n}\approx\mathrm{4}.\mathrm{28235} \\ $$
Commented by Tawa1 last updated on 10/Apr/19
Thank you sir.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by Tawa1 last updated on 10/Apr/19
If i really understand the spliting to  3^n  − 2^n . it will help.   Am sorry for disturbance sir
$$\mathrm{If}\:\mathrm{i}\:\mathrm{really}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{spliting}\:\mathrm{to}\:\:\mathrm{3}^{\mathrm{n}} \:−\:\mathrm{2}^{\mathrm{n}} .\:\mathrm{it}\:\mathrm{will}\:\mathrm{help}. \\ $$$$\:\mathrm{Am}\:\mathrm{sorry}\:\mathrm{for}\:\mathrm{disturbance}\:\mathrm{sir} \\ $$
Commented by Tawa1 last updated on 10/Apr/19
I don′t want to cram
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{want}\:\mathrm{to}\:\mathrm{cram} \\ $$
Commented by mr W last updated on 10/Apr/19
for how to solve equations like  3^n −2^n =65  there is only one way: that′s try and error.  since n≥1, you try with n=1,  3^1 −2^1 =1<65⇒not good, then try with n=2,  3^2 −2^2 =5<65⇒not good, then try with n=3,  3^3 −2^3 =19<65⇒not good, then try with n=4,  3^4 −2^4 =65=65⇒good, you got it!  to be sure you try with n=5,  3^5 −2^5 =211>65⇒not good, no other solution.
$${for}\:{how}\:{to}\:{solve}\:{equations}\:{like} \\ $$$$\mathrm{3}^{{n}} −\mathrm{2}^{{n}} =\mathrm{65} \\ $$$${there}\:{is}\:{only}\:{one}\:{way}:\:{that}'{s}\:{try}\:{and}\:{error}. \\ $$$${since}\:{n}\geqslant\mathrm{1},\:{you}\:{try}\:{with}\:{n}=\mathrm{1}, \\ $$$$\mathrm{3}^{\mathrm{1}} −\mathrm{2}^{\mathrm{1}} =\mathrm{1}<\mathrm{65}\Rightarrow{not}\:{good},\:{then}\:{try}\:{with}\:{n}=\mathrm{2}, \\ $$$$\mathrm{3}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} =\mathrm{5}<\mathrm{65}\Rightarrow{not}\:{good},\:{then}\:{try}\:{with}\:{n}=\mathrm{3}, \\ $$$$\mathrm{3}^{\mathrm{3}} −\mathrm{2}^{\mathrm{3}} =\mathrm{19}<\mathrm{65}\Rightarrow{not}\:{good},\:{then}\:{try}\:{with}\:{n}=\mathrm{4}, \\ $$$$\mathrm{3}^{\mathrm{4}} −\mathrm{2}^{\mathrm{4}} =\mathrm{65}=\mathrm{65}\Rightarrow{good},\:{you}\:{got}\:{it}! \\ $$$${to}\:{be}\:{sure}\:{you}\:{try}\:{with}\:{n}=\mathrm{5}, \\ $$$$\mathrm{3}^{\mathrm{5}} −\mathrm{2}^{\mathrm{5}} =\mathrm{211}>\mathrm{65}\Rightarrow{not}\:{good},\:{no}\:{other}\:{solution}. \\ $$
Commented by Tawa1 last updated on 10/Apr/19
Ohh, God bless you sir.  I appreciate your time sir.
$$\mathrm{Ohh},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}. \\ $$
Commented by Tawa1 last updated on 10/Apr/19
When you are chance sir,  i need the general term of  Σ_(i = 0) ^(n − 1)   ...
$$\mathrm{When}\:\mathrm{you}\:\mathrm{are}\:\mathrm{chance}\:\mathrm{sir},\:\:\mathrm{i}\:\mathrm{need}\:\mathrm{the}\:\mathrm{general}\:\mathrm{term}\:\mathrm{of}\:\:\underset{\mathrm{i}\:=\:\mathrm{0}} {\overset{\mathrm{n}\:−\:\mathrm{1}} {\sum}}\:\:… \\ $$
Commented by mr W last updated on 10/Apr/19
what′s your problem?  (1) you don′t understand how to get  3^n −2^n =65 from Σ_(i=0) ^(n − 1)     ^n C_i  2^i   =  65.  (2) you don′t understand how to get   n=4 from 3^n −2^n =65.
$${what}'{s}\:{your}\:{problem}? \\ $$$$\left(\mathrm{1}\right)\:{you}\:{don}'{t}\:{understand}\:{how}\:{to}\:{get} \\ $$$$\mathrm{3}^{{n}} −\mathrm{2}^{{n}} =\mathrm{65}\:{from}\:\underset{\mathrm{i}=\mathrm{0}} {\overset{\mathrm{n}\:−\:\mathrm{1}} {\sum}}\:\:\:\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{i}} \:\mathrm{2}^{\mathrm{i}} \:\:=\:\:\mathrm{65}. \\ $$$$\left(\mathrm{2}\right)\:{you}\:{don}'{t}\:{understand}\:{how}\:{to}\:{get} \\ $$$$\:{n}=\mathrm{4}\:{from}\:\mathrm{3}^{{n}} −\mathrm{2}^{{n}} =\mathrm{65}. \\ $$
Commented by Tawa1 last updated on 10/Apr/19
(1) is my problem sir.
$$\left(\mathrm{1}\right)\:\mathrm{is}\:\mathrm{my}\:\mathrm{problem}\:\mathrm{sir}.\:\: \\ $$
Commented by mr W last updated on 10/Apr/19
i can only give you examples, try to find  the rules by yourself:  a_1 +a_2 +a_3 +...+a_n =Σ_(i=1) ^n a_i   a_1 +a_2 +a_3 +...+a_n =a_1 +Σ_(i=2) ^n a_i   a_1 +a_2 +a_3 +...+a_(n−1) +a_n =Σ_(i=1) ^(n−1) a_i +a_n   a_1 +a_2 +a_3 +...+a_(n−2) +a_(n−1) +a_n =Σ_(i=2) ^(n−2) a_i +a_1 +a_(n−1) +a_n
$${i}\:{can}\:{only}\:{give}\:{you}\:{examples},\:{try}\:{to}\:{find} \\ $$$${the}\:{rules}\:{by}\:{yourself}: \\ $$$${a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +…+{a}_{{n}} =\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} \\ $$$${a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +…+{a}_{{n}} ={a}_{\mathrm{1}} +\underset{{i}=\mathrm{2}} {\overset{{n}} {\sum}}{a}_{{i}} \\ $$$${a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +…+{a}_{{n}−\mathrm{1}} +{a}_{{n}} =\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{a}_{{i}} +{a}_{{n}} \\ $$$${a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +…+{a}_{{n}−\mathrm{2}} +{a}_{{n}−\mathrm{1}} +{a}_{{n}} =\underset{{i}=\mathrm{2}} {\overset{{n}−\mathrm{2}} {\sum}}{a}_{{i}} +{a}_{\mathrm{1}} +{a}_{{n}−\mathrm{1}} +{a}_{{n}} \\ $$
Commented by mr W last updated on 10/Apr/19
3^n =(1+2)^n =Σ_(k=0) ^n C_k ^( n) 2^k =Σ_(k=0) ^(n−1) C_k ^( n) 2^k +C_n ^( n) 2^n   is this clear for you?
$$\mathrm{3}^{{n}} =\left(\mathrm{1}+\mathrm{2}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{\:{n}} \mathrm{2}^{{k}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{C}_{{k}} ^{\:{n}} \mathrm{2}^{{k}} +{C}_{{n}} ^{\:{n}} \mathrm{2}^{{n}} \\ $$$${is}\:{this}\:{clear}\:{for}\:{you}? \\ $$
Commented by Tawa1 last updated on 10/Apr/19
Yes sir, very clear sir. I realy appreciate your time sir. God bless you sir
$$\mathrm{Yes}\:\mathrm{sir},\:\mathrm{very}\:\mathrm{clear}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{realy}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$$$ \\ $$
Commented by mr W last updated on 10/Apr/19
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