Let (1+x)n=C0+C1x+C2x2+…+CnxnStatement-1: S=C0+C0+C1+C0+C1+C2+⋯+C0+C1+⋯+Cn−1=n2n−1Statement-2: ∑j=1n ∑i

# Let $\left(1+x{\right)}^{n}={C}_{0}+{C}_{1}x+{C}_{2}{x}^{2}+\dots +{C}_{n}{x}^{n}$Statement-1: $S={C}_{0}+\left({C}_{0}+{C}_{1}\right)+\left({C}_{0}+{C}_{1}+{C}_{2}\right)+\cdots +\left({C}_{0}+{C}_{1}+\cdots +{C}_{n-1}\right)=n\left({2}^{n-1}\right)$Statement-2: $\sum _{j=1}^{n} \sum _{i

1. A

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

2. B

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

3. C

STATEMENT-1 is True, STATEMENT-2 is False

4. D

STATEMENT-1 is False, STATEMENT-2 is True

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### Solution:

We can write

$S=n{C}_{0}+\left(n-1\right){C}_{1}+\left(n-2\right){C}_{2}+\cdots +1{C}_{n-1}+0{C}_{n}$

Using ${C}_{r}={C}_{n-r}$ we can rewrite (1) as

$S=0{C}_{0}+1{C}_{1}+2{C}_{2}+\cdots +\left(n-1\right){C}_{n-1}+n{C}_{n}$

Adding (1) and (2) we obtain

$2S=n\left[{C}_{0}+{C}_{1}+{C}_{2}+\cdots +{C}_{n}\right]=n\left({2}^{n}\right)$

In the  expression $\sum _{j=1}^{n} \sum _{i each ${C}_{i}\left(0\le i\le n\right)$occurs exactly n times. Thus

$\sum _{j=1}^{n} \sum _{i  