Let (1+x)n=C0+C1x+C2x2+…+Cnxn.Statement-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

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

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

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

Adding (1) and (2) we obtain

In the expression  $\sum _{j=1}^{n} \sum _{i each ${C}_{i}\left(O\le i\le n\right)\mathrm{Oc}-$ curs
exactly n times. Thus

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