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Question Number 115817 by A8;15: last updated on 28/Sep/20

Commented by Dwaipayan Shikari last updated on 28/Sep/20

$$\mathrm{Sorry},\mathrm{I}{\mathrm{haven}}^{\prime }\mathrm{t}\mathrm{found}\mathrm{any}\mathrm{symmetry}$$$$\mathrm{So}\mathrm{i}\mathrm{started}\mathrm{with}\mathrm{if}\mathrm{the}\mathrm{question}\mathrm{becomes}...$$

Commented by Dwaipayan Shikari last updated on 28/Sep/20

$$\mathrm{If}\mathrm{the}\mathrm{question}\mathrm{becomes}$$$$\mathrm{S}=\frac{1}{2}+\frac{3}{2^2}+\frac{6}{2^3}+\frac{10}{2^4}+\frac{15}{2^5}+..$$$$-\frac{\mathrm{S}}{2}=\frac{-1}{2^2}-\frac{3}{2^3}-\frac{6}{2^4}-..$$$$\frac{\mathrm{S}}{2}=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+....$$$${\mathrm{S}}^{\prime }=\frac{1}{2}+\frac{2}{2^2}+...({\mathrm{S}}^{\prime }=\frac{\mathrm{S}}{2})$$$$\frac{-{\mathrm{S}}^{\prime }}{2}=-\frac{1}{2^2}-.....$$$$\frac{{\mathrm{S}}^{\prime }}{2}=\frac{1}{2}+\frac{1}{2^2}+...=1\Rightarrow {\mathrm{S}}^{\prime }=2$$$$\mathrm{S}=4$$

Commented by A8;15: last updated on 28/Sep/20

this is a another solution

Commented by A8;15: last updated on 28/Sep/20

you send me solution of another question

Commented by MJS_new last updated on 28/Sep/20

$$\mathrm{the}\mathrm{answer}\mathrm{seems}\mathrm{to}\mathrm{be}6$$

Commented by prakash jain last updated on 28/Sep/20

$$S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{6}{2^4}+\frac{11}{2^5}+\frac{20}{2^6}+\frac{37}{2^7}$$$$\frac{S}{2}=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\frac{6}{2^5}$$$$\frac{S}{2^2}=\frac{1}{2^3}+\frac{2}{2^4}+\frac{3}{2^5}$$$$\frac{S}{2^3}=\frac{1}{2^4}+\frac{2}{2^5}$$$$S-\frac{S}{2}-\frac{S}{2^2}-\frac{S}{2^3}=\frac{1}{2}+\frac{1}{2^2}$$$$S(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8})=\frac{3}{4}$$$$S\left(\frac{1}{8}\right)=\frac{3}{4}$$$$S=6$$$$otheranswersarepossibleas$$$$generaltermformulafornumerator$$$$isnotgiven.$$

Commented by MJS_new last updated on 28/Sep/20

$$\mathrm{the}\mathrm{formula}\mathrm{seems}$$$$a_n=a_{n-3}+a_{n-2}+a_{n-1}\mathrm{with}{a}_1=1,a_2=2,a_3=3$$

Commented by prakash jain last updated on 28/Sep/20

$$\mathrm{Yes}.\mathrm{That}\mathrm{is}\mathrm{what}\mathrm{i}\mathrm{used}\mathrm{but}\mathrm{you},\mathrm{i}$$$$\mathrm{and}\mathrm{several}\mathrm{other}\mathrm{users}\mathrm{have}\mathrm{stated}$$$$\mathrm{if}\mathrm{a}\mathrm{finite}\mathrm{set}\mathrm{of}\mathrm{terms}\mathrm{are}\mathrm{given}\mathrm{then}$$$$\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{unique}\mathrm{formula}.$$$$a_1,a_2,....,a_k\mathrm{are}\mathrm{given}$$$$\mathrm{and}g\left(n\right)\mathrm{is}\mathrm{one}\mathrm{formuala}\mathrm{such}\mathrm{that}$$$$g\left(i\right)=a_i,1⩽i⩽k$$$$then$$$$f\left(n\right)=g\left(n\right)+h\left(n\right)\underset{i=1}{\prod ^k}(n-k)$$$$\mathrm{is}\mathrm{also}\mathrm{a}\mathrm{formula}\mathrm{where}h\left(n\right)\mathrm{is}$$$$\mathrm{any}\mathrm{arbitarary}\mathrm{function}\mathrm{of}n.$$

Commented by MJS_new last updated on 28/Sep/20

$$\mathrm{of}\mathrm{course}\mathrm{you}\mathrm{are}\mathrm{right}$$

Answered by Olaf last updated on 29/Sep/20

$$$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{a_n}{2^n}$$$$\mathrm{with}{a}_1=1,a_2=2,a_3=3$$$$\mathrm{and}{a}_n=a_{n-1}+a_{n-2}+a_{n-3},n⩾4$$$$\mathrm{S}=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\underset{n=4}{\sum ^{\mathrm{\infty }}}\frac{a_n}{2^n}$$$$\mathrm{S}=\frac{11}{8}+\underset{n=4}{\sum ^{\mathrm{\infty }}}\frac{a_{n-1}}{2^n}+\underset{n=4}{\sum ^{\mathrm{\infty }}}\frac{a_{n-2}}{2^n}+\underset{n=4}{\sum ^{\mathrm{\infty }}}\frac{a_{n-3}}{2^n}$$$$\mathrm{S}=\frac{11}{8}+\underset{n=3}{\sum ^{\mathrm{\infty }}}\frac{a_n}{2^{n+1}}+\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{a_n}{2^{n+2}}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{a_n}{2^{n+3}}$$$$\mathrm{S}=\frac{11}{8}+\frac{1}{2}\underset{n=3}{\sum ^{\mathrm{\infty }}}\frac{a_n}{2^n}+\frac{1}{4}\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{a_n}{2^n}+\frac{1}{8}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{a_n}{2^n}$$$$\mathrm{S}=\frac{11}{8}+\frac{1}{2}(\mathrm{S}-\frac{1}{2}-\frac{2}{4})+\frac{1}{4}(\mathrm{S}-\frac{1}{2})+\frac{1}{8}\mathrm{S}$$$$\mathrm{S}(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8})=\frac{11}{8}-\frac{1}{2}-\frac{1}{8}$$$$\frac{1}{8}\mathrm{S}=\frac{6}{8}$$$$\mathrm{S}=6$$

Commented by prakash jain last updated on 29/Sep/20

$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}=\frac{4+4+3}{8}=\frac{11}{8}$$

Commented by Olaf last updated on 29/Sep/20

$$\mathrm{Yes}!\mathrm{Thank}\mathrm{you}\mathrm{mister}.\mathrm{I}\mathrm{corrected}.$$

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