Let(1+t)n=C0+C1t+C2t2+⋯+CntnStatement-1: C0(1)(2)+C1(2)(3)+C2(3)(4)+⋯+Cn(n+1)(n+2)                    =1n+12n+2n+2−1Statement-2:  C02−C13+C24−⋯+(−1)nCnn+2=0

# Let$\left(1+t{\right)}^{n}={C}_{0}+{C}_{1}t+{C}_{2}{t}^{2}+\cdots +{C}_{n}{t}^{n}$Statement-1: $\frac{{C}_{0}}{\left(1\right)\left(2\right)}+\frac{{C}_{1}}{\left(2\right)\left(3\right)}+\frac{{C}_{2}}{\left(3\right)\left(4\right)}+\cdots +\frac{{C}_{n}}{\left(n+1\right)\left(n+2\right)}$                    $=\frac{1}{n+1}\left[\frac{{2}^{n+2}}{n+2}-1\right]$Statement-2:  $\frac{{C}_{0}}{2}-\frac{{C}_{1}}{3}+\frac{{C}_{2}}{4}-\cdots +\left(-1{\right)}^{n}\frac{{C}_{n}}{n+2}=0$

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:

From (1)

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

$=\frac{{C}_{0}}{1}x+\frac{{C}_{2}}{2}{x}^{2}+\cdots +\frac{{C}_{n}}{n+1}{x}^{n+1}$                           (2)

Multiplying (1) by t and integrating, we get

$=\frac{{C}_{0}}{2}{x}^{2}+\frac{{C}_{1}}{3}{x}^{3}+\frac{{C}_{2}}{4}{x}^{4}+\dots +\frac{{C}_{n}}{n+2}{x}^{n+2}.$            (3)

Putting in (3), we obtain that the second statement is true

Putting in (2) and (3) and subtracting, we obtain that Statement-1 is also true.

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