KVPY 2020-21 SB SX Exam Previous Papers

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Question : 81 of 120

Marks: +1, -0

PART-II : MATHEMATICS

Let

a=\∑↙{n=101}↖{200} 2^{n} \∑↙{k=101}↖{n} {1}/{k !}

and

b=\∑↙{n=101}↖{200} {2^{201}-2^{n}}/{n !}

1
$3/2$
2
$5/2$

Solution:

a=\∑↙{n=101}↖{200} 2^{n} \∑↙{k=101}↖{n} {1}/{k !}

={2^{101}}/{101 !}+2^{102}({1}/{101 !}+{1}/{102 !})+2^{103}({1}/{101 !}

+{1}/{102 !}+{1}/{103 !})+\......

+2^{200}({1}/{101 !}+{1}/{102 !}+\....+{1}/{200 !})

{2^{101}+\.....+2^{200}}/{101 !} +{2^{102} +\...... +2^{200}}/{102 !}

+\......+{2^{200}}/{200 !}

={2^{101}(2^{100}-1)}/{101 !}+{2^{102}(2^{99}-1)}/{102 !}

+\.....+{2^{200}}/{200 !}

=({2^{201}}/{101 !}-{2^{101}}/{101 !})+({2^{201}}/{102 !}-{2^{102}}/{102 !})

+\.......+({2^{201}}/{200 !}-{2^{200}}/{200 !})

=∑↙{n=101}↖{200} {2^{201}-2^{n}}/{n !}=b

\∴{a}/{b}=1

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