If {(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} | vedclass

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(b) ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ..... + {C_n}{x^n}$ .....$(i)$

and ${\left( {1 + \frac{1}{x}} \right)^n} = {C_0} + {C_1}\frac{1}{x

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ..... + {C_n}{x^n}$ .....$(i)$

and ${\left( {1 + \frac{1}{x}} ight)^n} = {C_0} + {C_1}\frac{1}{x} + {C_2}{\left( {\frac{1}{x}} ight)^2} + ..... + {C_n}{\left( {\frac{1}{x}} ight)^n}$ ....$(ii)$ 

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 ${(1 + x)^n}{\left( {1 + \frac{1}{x}} ight)^n}$

or in $\frac{1}{{{x^n}}}{(1 + x)^{2n}}$ or term containing ${x^n}$ in ${(1 + x)^{2n}}$. 

Clearly the coefficient of ${x^n}$ in ${(1 + x)^{2n}}$is ${T_{n + 1}}$ and equal to $^{2n}{C_n} = \frac{{(2n)!}}{{n!\,\,n!}}$

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 $\frac{{4!}}{{2!\,2!}} = \frac{{4 	imes 3 	imes 2 	imes 1}}{{2 	imes 1 	imes 2 	imes 1}} = 6$

Note : Students should remember this question as an identity.

If ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .......... + {C_n}{x^2},$ then $C_0^2 + C_1^2 + C_2^2 + C_3^2 + ...... + C_n^2$ =

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