"Let $\left\{ {{a_n}} \right\}_{n = 1}^\infty$ be a sequence such that a1 = 1, a2 = 1 and ${a_{n + 2}} = 2{a_{n + 1}} + {a_n}$ for all n $\ge$ 1. Then the value of $47\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{2^{3n}}}}}$ is equal to ______________."

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Accepted Answer

"${a_{n + 2}} = 2{a_{n + 1}} + {a_n}$, let $\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{8^n}}}} = P$

Divide by 8ⁿ we get

$${{{a_{n + 2}}} \over {{8^n}}} = {{2{a_{n + 1}}} \over {{8^n}}} + {{{a_n}} \over {{8^n}}}$$

$$ \Rightarrow 64{{{a_{n + 2}}} \over {{8^{n + 2}}}} = {{16{a_{n + 1}}} \over {{8^{n + 1}}}} + {{{a_n}} \over {{8^n}}}$$

$$64\sum\limits_{n = 1}^\infty {{{{a_{n + 2}}} \over {{8^{n + 2}}}}} = 16\sum\limits_{n = 1}^\infty {{{{a_{n + 1}}} \over {{8^{n + 1}}}}} + \sum\limits_{n = 1}^\infty {{{{a_n}} \over {{8^n}}}} $$

$$64\left( {P - {{{a_1}} \over 8} - {{{a_2}} \over {{8^2}}}} \right) = 16\left( {P - {{{a_1}} \over 8}} \right) + P$$

$$ \Rightarrow 64\left( {P - {1 \over 8} - {1 \over {64}}} \right) = 16\left( {P - {1 \over 8}} \right) + P$$

$64P - 8 - 1 = 16P - 2 + P$

$47P = 7$"

Let $\left\{ {{a_n}} \right\}_{n = 1}^\infty$ be a sequence such that a₁ = 1, a₂ = 1 and ${a_{n + 2}} = 2{a_{n + 1}} + {a_n}$ for all n $\ge$ 1. Then the value of $47\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{2^{3n}}}}}$ is equal to ______________.

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Let $\left\{ {{a_n}} \right\}_{n = 1}^\infty$ be a sequence such that a1 = 1, a2 = 1 and ${a_{n + 2}} = 2{a_{n + 1}} + {a_n}$ for all n $\ge$ 1. Then the value of $47\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{2^{3n}}}}}$ is equal to ______________.

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Let $\left\{ {{a_n}} \right\}_{n = 1}^\infty$ be a sequence such that a₁ = 1, a₂ = 1 and ${a_{n + 2}} = 2{a_{n + 1}} + {a_n}$ for all n $\ge$ 1. Then the value of $47\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{2^{3n}}}}}$ is equal to ______________.