Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

Question Number 57688 by Tawa1 last updated on 10/Apr/19

$$\mathrm{Solve}\mathrm{for}\mathrm{n}:\underset{\mathrm{i}}{\sum ^{\mathrm{n}-1}}\stackrel{\mathrm{n}}{}{\mathrm{C}}_{\mathrm{i}}{2}^{\mathrm{i}}=65,\mathrm{n}\in {\mathbb{Z}}^{+}.\mathrm{where}\mathrm{zero}\mathrm{is}$$$$\mathrm{included}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Apr/19

$$questionnotclear..$$$$\underset{i=0}{\sum ^{n-1}}{C}_i^n{2}^i\leftarrow isitthequestion$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{Yes}\mathrm{sir}.$$

Commented by Abdo msup. last updated on 10/Apr/19

$$firstwehave\sum _{i=0}^{n-1}{C}_n^i{2}^i=\sum _{i=0}^n{C}_n^i{2}^i-C_n^n{2}^n$$$$=3^n-2^n\left(wesupposethatthesumbeginsfrom0\right)$$$$\Rightarrow \left(e\right)\Rightarrow 3^n-2^n=65\Rightarrow n=4.$$$$$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sirs},\mathrm{is}\mathrm{there}\mathrm{anyway}\mathrm{to}\mathrm{find}\mathrm{n}\mathrm{from}{3}^{\mathrm{n}}-2^{\mathrm{n}}=65$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{Sir},\mathrm{please}\mathrm{i}\mathrm{need}\mathrm{the}\mathrm{rules}\mathrm{of}\mathrm{summation},\mathrm{expecially}\mathrm{the}\mathrm{sum}$$$$\mathrm{that}\mathrm{has}\mathrm{something}\mathrm{like}\underset{\mathrm{i}=0}{\sum ^{\mathrm{n}-1}}\mathrm{how}\mathrm{to}\mathrm{split}\mathrm{them}$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{Because}\mathrm{have}\mathrm{been}\mathrm{trying}\mathrm{to}\mathrm{understand}\mathrm{how}\mathrm{to}\mathrm{get}$$$$3^{\mathrm{n}}-2^{\mathrm{n}}=65.\mathrm{I}\mathrm{did}\mathrm{not}\mathrm{understand}\mathrm{yet}.$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{Sorry}\mathrm{for}\mathrm{disturbing}\mathrm{sir}$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{or}\mathrm{textbook}\mathrm{i}\mathrm{can}\mathrm{learn}\mathrm{it}.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Apr/19

$$S=C_0^n{2}^0+C_1^n{2}^1+C_2^n{2}^2+....+C_{n-1}^n{2}^{n-1}$$$$nowlookhere$$$${(x+2)}^n=C_0^n{x}^{n-0}{2}^0+C_1^n{x}^{n-1}{2}^1+C_2^n{x}^{n-2}{2}^2+...+C_{n-1}^n{x}^1{2}^{n-1}+C_n^n{x}^{n-n}{2}^n$$$$nowputx=1bothside$$$$3^n=[C_0^n{2}^0+C_1^n{2}^1+...+C_{n-1}^n{2}^{n-1}]+C_n^n{2}^n\leftarrow lookhere$$$$3^n=S+C_n^n{2}^n$$$$now{C}_n^n=\frac{n!}{n!(n-n)!}=1$$$$so$$$$3^n=S+2^n\times 1$$$$S=3^n-2^n$$$$so\underset{i=0}{\sum ^{n-1}}{C}_i^{n_{}}{2}^i=\mathit{S}=3^n-2^n$$$$3^n-2^n=65$$$$bylnspection$$$$3^n-2^n=81-16$$$$3^n-2^n=3^4-2^4\to n=4$$$$$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{Ohh},\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}\mathrm{and}\mathrm{effort}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{I}\mathrm{really}\mathrm{appreciate}\mathrm{sir}$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{And}\mathrm{i}\mathrm{understand}\mathrm{very}\mathrm{well}$$

Answered by mr W last updated on 10/Apr/19

$$assumedyoumean$$$$\underset{\mathrm{i}=0}{\sum ^{\mathrm{n}-1}}\stackrel{\mathrm{n}}{}{\mathrm{C}}_{\mathrm{i}}{2}^{\mathrm{i}}=65$$$$LHS={(1+2)}^n-2^n$$$${(1+2)}^n-2^n=65$$$$3^n-2^n=65$$$$\Rightarrow n=4$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{How}\mathrm{is}\mathrm{left}\mathrm{hand}\mathrm{side}={(1+2)}^{\mathrm{n}}-2^{\mathrm{n}}\mathrm{sir}??$$

Commented by mr W last updated on 10/Apr/19

$${(1+2)}^n=\underset{i=0}{\sum ^n}{C}_i^n{2}^i=\underset{i=0}{\sum ^{n-1}}{C}_i^n{2}^i+2^n=LHS+2^n$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{any}\mathrm{step}\mathrm{to}\mathrm{get}\mathrm{n}=4?$$

Commented by mr W last updated on 10/Apr/19

$$tryandsee...$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{Trier}\mathrm{and}\mathrm{error}?$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{I}\mathrm{found}\mathrm{a}\mathrm{method}\mathrm{i}\mathrm{used}\mathrm{to}\mathrm{get}\mathrm{n}=4,\mathrm{from}{3}^{\mathrm{n}}-2^{\mathrm{n}}=65.\mathrm{I}\mathrm{used}\mathrm{algebra}$$$$\mathrm{representation}.$$$$\mathrm{But}\mathrm{am}\mathrm{still}\mathrm{finding}\mathrm{it}\mathrm{difficult}\mathrm{to}\mathrm{understand}.\mathrm{how}\mathrm{to}\mathrm{get}$$$$3^{\mathrm{n}}-2^{\mathrm{n}}=65\mathrm{from}\mathrm{the}\mathrm{relation}.$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{Maybe}\mathrm{i}\mathrm{need}\mathrm{sigma}\mathrm{notation}\mathrm{rules}.\mathrm{How}\mathrm{to}\mathrm{apply}$$

Commented by Tawa1 last updated on 10/Apr/19

$$3^{\mathrm{n}}-2^{\mathrm{n}}=65$$$${\left[3^{\left(\mathrm{n}/2\right)}\right]}^2-{\left[2^{\left(\mathrm{n}/2\right)}\right]}^2=65$$$$3^{\mathrm{n}/2}=\mathrm{m}\mathrm{and}{2}^{\mathrm{n}/2}=\mathrm{y}$$$${\mathrm{m}}^2-{\mathrm{y}}^2=65$$$$(\mathrm{m}+\mathrm{y})(\mathrm{m}-\mathrm{y})=13\times 5$$$$\mathrm{m}+\mathrm{y}=13$$$$\mathrm{m}-\mathrm{y}=5$$$$\mathrm{m}=9,\mathrm{y}=4$$$$3^{\mathrm{n}/2}=3^2\mathrm{and}{2}^{\mathrm{n}/2}=2^2$$$$\mathrm{n}=4\mathrm{twice}.$$$$$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{But}\mathrm{am}\mathrm{still}\mathrm{comfused}\mathrm{on}\mathrm{how}\mathrm{to}\mathrm{get}{3}^{\mathrm{n}}-2^{\mathrm{n}}\mathrm{from}\mathrm{the}\mathrm{sum}$$

Commented by MJS last updated on 10/Apr/19

$$\mathrm{this}{\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{a}\mathrm{method},\mathrm{it}\mathrm{works}\mathrm{here}\mathrm{just}\mathrm{by}\mathrm{chance}$$$$3^n-2^n=91$$$${\left(3^{n/2}\right)}^2-{\left(2^{n/2}\right)}^2=91$$$$(3^{n/2}-2^{n/2})(3^{n/2}+2^{n/2})=7\times 13$$$$3^{n/2}-2^{n/2}=7$$$$3^{n/2}+2^{n/2}=13$$$$\Rightarrow 3^{n/2}=10\wedge 2^{n/2}=3$$$$\Rightarrow n=\frac{2\ln 10}{\ln 3}\wedge n=\frac{2\ln 3}{\ln 2}$$$$\mathrm{while}\mathrm{in}\mathrm{this}\mathrm{case}n\approx 4.28235$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}.$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{If}\mathrm{i}\mathrm{really}\mathrm{understand}\mathrm{the}\mathrm{spliting}\mathrm{to}{3}^{\mathrm{n}}-2^{\mathrm{n}}.\mathrm{it}\mathrm{will}\mathrm{help}.$$$$\mathrm{Am}\mathrm{sorry}\mathrm{for}\mathrm{disturbance}\mathrm{sir}$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{want}\mathrm{to}\mathrm{cram}$$

Commented by mr W last updated on 10/Apr/19

$$forhowtosolveequationslike$$$$3^n-2^n=65$$$$thereisonlyoneway:{that}^{\prime }stryanderror.$$$$sincen⩾1,youtrywithn=1,$$$$3^1-2^1=1<65\Rightarrow notgood,thentrywithn=2,$$$$3^2-2^2=5<65\Rightarrow notgood,thentrywithn=3,$$$$3^3-2^3=19<65\Rightarrow notgood,thentrywithn=4,$$$$3^4-2^4=65=65\Rightarrow good,yougotit!$$$$tobesureyoutrywithn=5,$$$$3^5-2^5=211>65\Rightarrow notgood,noothersolution.$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{Ohh},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}\mathrm{sir}.$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{When}\mathrm{you}\mathrm{are}\mathrm{chance}\mathrm{sir},\mathrm{i}\mathrm{need}\mathrm{the}\mathrm{general}\mathrm{term}\mathrm{of}\underset{\mathrm{i}=0}{\sum ^{\mathrm{n}-1}}...$$

Commented by mr W last updated on 10/Apr/19

$${what}^{\prime }syourproblem?$$$$\left(1\right)you{don}^{\prime }tunderstandhowtoget$$$$3^n-2^n=65from\underset{\mathrm{i}=0}{\sum ^{\mathrm{n}-1}}\stackrel{\mathrm{n}}{}{\mathrm{C}}_{\mathrm{i}}{2}^{\mathrm{i}}=65.$$$$\left(2\right)you{don}^{\prime }tunderstandhowtoget$$$$n=4from{3}^n-2^n=65.$$

Commented by Tawa1 last updated on 10/Apr/19

$$\left(1\right)\mathrm{is}\mathrm{my}\mathrm{problem}\mathrm{sir}.$$

Commented by mr W last updated on 10/Apr/19

$$icanonlygiveyouexamples,trytofind$$$$therulesbyyourself:$$$$a_1+a_2+a_3+...+a_n=\underset{i=1}{\sum ^n}{a}_i$$$$a_1+a_2+a_3+...+a_n=a_1+\underset{i=2}{\sum ^n}{a}_i$$$$a_1+a_2+a_3+...+a_{n-1}+a_n=\underset{i=1}{\sum ^{n-1}}{a}_i+a_n$$$$a_1+a_2+a_3+...+a_{n-2}+a_{n-1}+a_n=\underset{i=2}{\sum ^{n-2}}{a}_i+a_1+a_{n-1}+a_n$$

Commented by mr W last updated on 10/Apr/19

$$3^n={(1+2)}^n=\underset{k=0}{\sum ^n}{C}_k^n{2}^k=\underset{k=0}{\sum ^{n-1}}{C}_k^n{2}^k+C_n^n{2}^n$$$$isthisclearforyou?$$

Commented by Tawa1 last updated on 10/Apr/19

$$\mathrm{Yes}\mathrm{sir},\mathrm{very}\mathrm{clear}\mathrm{sir}.\mathrm{I}\mathrm{realy}\mathrm{appreciate}\mathrm{your}\mathrm{time}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$$$$$

Commented by mr W last updated on 10/Apr/19

��

Terms of Service

Privacy Policy

Contact: info@tinkutara.com