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Let \mathrm{\left\{a_{n}\right\}_{n=0}^{\infty}}  be a sequence such that \mathrm{a_{0}=a_{1}=0 \: and \: a_{n+2}=2 a_{n+1}-a_{n}+1} for all n \geqslant 0.
Then, \sum_{n=2}^{\infty} \frac{a_{n}}{7^{n}} is equal to:

Option: 1

\frac{4}{343}


Option: 2

\frac{7}{216}


Option: 3

\frac{8}{343}


Option: 4

\frac{49}{216}


Answers (1)

best_answer

\mathrm{a_{0}=a_{1}=0 } \\

\mathrm{a_{n+2}=2 a_{n+1}-a_{n}+1} \\

\mathrm{\Rightarrow \frac{a_{n+2}}{7^{n+2}}=2 \frac{a_{n+1}}{7^{n+2}}-\frac{a_{n}}{7^{n+2}}+\frac{1}{7^{n+2}} }

\mathrm{\Rightarrow \frac{a_{n+2}}{7^{n+2}}=\frac{2}{7} \cdot\left(\frac{a_{n+1}}{7^{n+1}}\right)-\frac{1}{7^{2}}\left(\frac{a_{n}}{7^{n}}\right)+\frac{1}{7^{n+2}} }

Applying summation from \mathrm{n=0 } to infinity

\mathrm{\Rightarrow \sum_{n=0}^{\infty} \frac{a_{n+2}}{7^{n+2}}=\frac{2}{7} \sum_{0}^{\infty} \frac{a_{n+1}}{7^{n+1}}-\frac{1}{49} \sum_{0}^{\infty} \frac{a_{n}}{7^{n}} +\sum_{0}^{\infty} \frac{1}{7^{n+2}}}       ......(i)

Now

\mathrm{\sum_{0}^{\infty} \frac{a_{n+2}}{7^{n+2}}=\frac{a_{2}}{7^{2}}+\frac{a_{3}}{7^{3}}+\cdots=S \text { (let) }}

And   \mathrm{\sum_{0}^{\infty} \frac{a_{n+1}}{7^{n+1}}=\frac{a_{1}}{7}+\frac{a_{2}}{7^{2}}+\frac{a_{3}}{7^{3}}+\cdots}

                              \mathrm{=\frac{a_{2}}{7^{2}}+\frac{a_{3}}{7^{3}}+\cdots \cdot\left(\operatorname{as} a_{1}=0\right)} \\

                              \mathrm{=S}

And  \mathrm{\sum_{0}^{\infty} \frac{a_{n}}{7^{n}} =\frac{a_{0}}{1}+\frac{a_{1}}{7}+\frac{a_{2}}{7^{2}}+\cdots} \\

                          \mathrm{=\frac{a_{2}}{7^{2}}+\frac{a_{3}}{7^{3}}+\cdots=S }

From (i)

\mathrm{\Rightarrow S=\frac{2}{7} S-\frac{1}{49} S+\left(\frac{1}{7^{2}}+\frac{1}{7^{3}}+\cdots\right)} \\

\mathrm{\Rightarrow S-\frac{2}{7} S+\frac{1}{49} S=\frac{1}{49} \cdot \frac{1}{1-\frac{1}{7}}} \\

\mathrm{\Rightarrow \frac{(49-14+1) S}{49}=\frac{1}{7 \cdot 6} }

\mathrm{\Rightarrow S=\frac{49}{7\cdot 6\cdot 36}}

\mathrm{\Rightarrow S=\frac{7}{216}}

Hence the correct answer is option 2

Posted by

Deependra Verma

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