First slide

Binomial theorem for positive integral Index

Question

$If (1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ then $C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + C_{3}^{2} + \dots + C_{n}^{2}$ is equal to

Moderate

A

$\frac{n!}{n! n!}$
B

$\frac{(2 n)!}{n! n!}$
C

$\frac{(2 n)!}{n!}$
D

None of these

Solution

we have , $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n} - - - - - (i)$
$and {(1 + \frac{1}{x})}^{n} = C_{0} + C_{1} \frac{1}{x} + C_{2} {(\frac{1}{x})}^{2} + \dots + C_{n} {(\frac{1}{x})}^{n} \dots (ii)$
on multiplying Eqs. (i) and (ii) and taking the
coefficient of constant terms in right hand side
$= C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + \dots + C_{n}^{2}$
In right hand side $(1 + x)^{n} {(1 + \frac{1}{x})}^{n} or in \frac{1}{x^{n}} (1 + x)^{2 n} or$
term containing xⁿ in (1 + x)²ⁿ.
clearly, the coefficient of xⁿ in (1 + x)²ⁿ is equal to
$^{2 n} C_{n} = \frac{(2 n)!}{n! n!}$

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