Let $\{ {a_n}\} _{n = 0}^\infty$ be a sequence such that ${a_0} = {a_1} = 0$ and ${a_{n + 2}} = 2{a_{n + 1}} - {a_n} + 1$ for all n $\ge$ 0. Then, $\sum\limits_{n = 2}^\infty {{{{a_n}} \over {{7^n}}}}$ is equal to:
Solution
<p>${a_{n + 2}} = 2{a_{n + 1}} - {a_n} + 1$ & ${a_0} = {a_1} = 0$</p>
<p>${a_2} = 2{a_1} - {a_0} + 1 = 1$</p>
<p>${a_3} = 2{a_2} - {a_1} + 1 = 3$</p>
<p>${a_4} = 2{a_3} - {a_2} + 1 = 6$</p>
<p>${a_5} = 2{a_4} - {a_3} + 1 = 10$</p>
<p>$$\sum\limits_{n = 2}^\infty {{{{a_n}} \over {{7^n}}} = {{{a_2}} \over {{7^2}}} + {{{a_3}} \over {{7^3}}} + {{{a_4}} \over {{7^4}}} + \,\,...} $$</p>
<p>$$s = {1 \over {{7^2}}} + {3 \over {{7^3}}} + {6 \over {{7^4}}} + {{10} \over {{7^5}}} + \,\,...$$</p>
<p>$${{{1 \over 7}s = {1 \over {{7^3}}} + {3 \over {{7^4}}} + {6 \over {{7^5}}} + \,\,...} \over {{{6s} \over 7} = {1 \over {{7^2}}} + {2 \over {{7^3}}} + {3 \over {{7^4}}} + \,\,...}}$$</p>
<p>$${{{{6s} \over {49}} = \,\,\,\,\,\,\,\,\,{1 \over {{7^3}}} + {2 \over {{7^4}}} + \,\,...} \over {{{36s} \over {49}} = {1 \over {{7^2}}} + {1 \over {{7^3}}} + {1 \over {{7^4}}} + \,\,...}}$$</p>
<p>${{36s} \over {49}} = {{{1 \over {{7^2}}}} \over {1 - {1 \over 7}}}$</p>
<p>${{36s} \over {49}} = {7 \over {49 \times 6}}$</p>
<p>$s = {7 \over {216}}$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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