If ( 1 + x ) n = C 0 + C 1 x + C 2 x 2 + . . . . . . . . + C n x 2 , then C 2 0 + C 2 1 + C 2 2 + C 2 3 + . . . . . . . . . . + C 2 n = Options: A . &nbsp n ! n ! n ! B . &nbsp ( 2 n ) ! n ! n ! C . &nbsp ( 2 n ) ! n ! D . &nbsp None of these

If (1+x)n=C0+C1x+C2x2+........+Cnx2, then C20+C21+C22+C23+..........+C2

Discussion Forum : Binomial Theorem

Question -

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . . + C_{n} x^{2}$ , then

$C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + C_{3}^{2} + . . . . . . . . . . + C_{n}^{2}$ =

Options:

A .

\frac{n!}{n! n!}

B .

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

C .

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

D . None of these

Answer: Option B
:
B

$(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . . . + C_{n} X^{n}$ ...........(i)

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

If we multiply (i) and (ii), we get

$C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + . . . . . . . . + C_{n}^{2}$

is the term independent of x and hence it is equal to the term independent of x in the product $(1 + x)^{n} (1 + \frac{1}{x})^{n}$ or in $\frac{1}{x^{n}}$ $(1 + x)^{2 n}$ or term

containing $x^{n}$ in $(1 + x)^{2 n}$ . Clearly the coefficient of $x^{n}$ in $(1 + x)^{2 x}$ is $T_{n + 1}$ and equal to $^{2 n} C_{n}$ = $\frac{(2 n)!}{n! n!}$

Trick: Solving conversely.

Put n = 1, n = 2,.........then we get $S_{1}$ = $^{1} C_{0}^{2} +^{1} C_{1}^{2}$ = 2,

$S_{2}$ = $^{2} C_{0}^{2} +^{2} C_{1}^{2} +^{2} C_{2}^{2} = 1^{2} + 2^{2} + 1^{2}$ = 6

Now check the options
(a) Does not hold given condition,
(b) (i) Put n = 1, then $\frac{2!}{1! 1!} = 2$
(ii) Put n = 2, then $\frac{4!}{2! 2!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1} = 6$
Note: Students should remember this question as identity

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