KVPY SX SB Exam 19-Nov-2017 Question Paper

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Question : 4 of 120

Marks: +1, -0

Let S be the infinite sum given by

S=∑↖{∞}↙{n=0}a_n/10^{2n}

where

\{an\}_{n≥0}

is a sequence defined by

a_0 = a_1 = 1

and

a_j = 20a_{j-1} -108 a_{j–2} \for j ≥2.

If S is expressed in the form

a/b

, where a, b are coprime positive integers, then a equals

2017
2020
2023
2025

Solution:

a_n = 20a_{n–1} - 108a_{n–2}

a_n/10^{2n}={20a_{n–1}}/10^{2n}-{108a_{n–2}}/10^{2n}

a_n/10^{2n}=20/100\ a _{n-1}/ 10^{2(n-1)}-108/10000 \ a _{n-2}/ 10^{2(n-2)}

apply summation

∑↖{∞}↙{n=0}a_n/10^{2n}=1/5 ∑↖{∞}↙{n=0}a_{n-1}/10^{2(n-1)}

-27/2500 ∑↖{∞}↙{n=0}a_{n-2}/10^{2(n-2)}

S - 1 -1/100=1/5(S- 1) -27/2500\ S.

S - 1 -1/100=1/5\ S- 1/5 -27/2500\ S.

S(1-1/5+27/2500)=-1/5+1+1/100

S({2500-500+27}/2500)=81/100

S={81×25}/2027

S=2025/2027

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