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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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