KVPY SX SB Exam 19-Nov-2017 Question Paper

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Question : 15 of 120

Marks: +1, -0

Let

I_n = ∫ ↖{1} ↙{0} e^{-y} y ^{n}dy

, where n is a non-negative integer. Then

∑↖{∞}↙{n=1}I_n /n!

1
$1- 1 /e$
$1/e$
$1+ 1/e$

Solution:

I_n = ∫ ↖{1} ↙{0} e^{-y} y ^{n}dy

I_n =-1/e+n\ I_{n-1}

(by reduction formula)

I_n -n \ I_{n-1} =-1/e.

I_n/n!-I_{n-1}/{n-1!}=-I/{e(n!)

n = 1

\;\;\;\;I_1/1!-I_0/0!=-1/e

...(i)
n = 2

\;\;\;\;I_2/2!-I_1/1!=-1/{e(2!)}

...(ii)

n = 3

\;\;\;\;I_3/3!-I_2/2!=-1/{e(3!)}

...(iii)

n = 4

\;\;\;\;I_4/4!-I_3/3!=-1/{e(4!)}

...(iii)
eq.(1) + eq.(2) × 2 + eq.(3) × 3 + (eq. 4) × 4 + ......

-\{I_1/1!+I_2/2!+I_3/3!+I_4/4!\}-I_0

=-1/e\{1+1+1/2!+1/3!+...\}

Let S =

∑↖{∞}↙{n=1}I_n /n!

-S -I_0 = -1/e×e

S = 1 -I_0

S = 1 -∫ ↖{1} ↙{0} e^{-y} dy

S = 1 +[e^{-y}]^1_0

S = 1 +[1/e-1]

S = 1 /e

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