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

A
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
B
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
C
STATEMENT-1 is True, STATEMENT-2 is False
D
STATEMENT-1 is False, STATEMENT-2 is True

Solution:

From (1)

$\int_{0}^{x} (1 + t)^{n} d t = \int_{0}^{x} [C_{0} + C_{1} t + C_{2} t^{2} + \dots + C_{n} t^{n}] d t$

$\Rightarrow \frac{1}{n + 1} [(1 + x)^{n + 1} - 1]$

$= \frac{C_{0}}{1} x + \frac{C_{2}}{2} x^{2} + \dots + \frac{C_{n}}{n + 1} x^{n + 1}$ (2)

Multiplying (1) by t and integrating, we get

$\begin{array}{l} \int_{0}^{x} t (1 + t)^{n} d t = \int_{0}^{x} [C_{0} t + C_{1} t^{2} + \dots + C_{n} t^{n + 1}] d t \\ \Rightarrow \frac{x (1 + x)^{n + 1}}{n + 1} - \frac{(1 + x)^{n + 2}}{(n + 1) (n + 2)} \end{array}$

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

Putting $x = - 1$ in (3), we obtain that the second statement is true

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

Solution:

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