Let S be the infinite sum given by S = sum_{n | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The sequence is defined by $a_n = 20a_{n-1} - 108a_{n-2}$ for $n \geq 2$ with $a_0 = 1, a_1 = 1$.Multiply by $\frac{1}{10^{2n}}$ and sum from $n=2

Vedclass · Accepted Answer

$2025$

Vedclass · Answer

(D) The sequence is defined by $a_n = 20a_{n-1} - 108a_{n-2}$ for $n \geq 2$ with $a_0 = 1, a_1 = 1$.Multiply by $\frac{1}{10^{2n}}$ and sum from $n=2$ to $\infty$:$\sum_{n=2}^{\infty} \frac{a_n}{10^{2n}} = \sum_{n=2}^{\infty} \frac{20a_{n-1}}{10^{2n}} - \sum_{n=2}^{\infty} \frac{108a_{n-2}}{10^{2n}}$$S - a_0 - \frac{a_1}{100} = \frac{20}{100} \sum_{n=2}^{\infty} \frac{a_{n-1}}{10^{2(n-1)}} - \frac{108}{10000} \sum_{n=2}^{\infty} \frac{a_{n-2}}{10^{2(n-2)}}$$S - 1 - \frac{1}{100} = \frac{1}{5} (S - a_0) - \frac{108}{10000} S$$S - \frac{101}{100} = \frac{1}{5} (S - 1) - \frac{27}{2500} S$$S - \frac{1}{5} S + \frac{27}{2500} S = 1 + \frac{1}{100} - \frac{1}{5}$$S \left( \frac{2500 - 500 + 27}{2500} ight) = \frac{100 + 1 - 20}{100}$$S \left( \frac{2027}{2500} ight) = \frac{81}{100}$$S = \frac{81 	imes 25}{2027} = \frac{2025}{2027}$Since $a = 2025$ and $b = 2027$ are coprime,$a = 2025$.

Similar Questions

Find the sum to $n$ terms of the series whose $n^{th}$ term is given by $a_n = n(n+1)(n+4)$.

$\sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^{n - 1} {\frac{k}{{{2^{n + k}}}}} } $ is equal to

The value of $\sum\limits_{r = 16}^{30} {(r + 2)(r - 3)}$ is equal to

The expression for $a_n$ which satisfies $a_0=0, a_1=1$ and $a_n=a_{n-1}+a_{n-2}, \forall n \in N -\{0,1\}$ is:

The sum of the series $3.6 + 4.7 + 5.8 + \dots$ up to $(n - 2)$ terms is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module