AP EAPCET 15th May 2023 Shift 1 Solved Paper

Section: Mathematics

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Question : 14 of 160

Marks: +1, -0

If Cj stands for ‌nCj, then ‌

C1

C0

+‌

2×C2

C1

+‌

3×C3

C2

+...+‌

n×Cn

Cn−1

=

n
∑
k=1
k2
n
∑
k=1
‌
k
2
n
∑
k=1
2k
n
∑
k=1
k

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