If (1 + x)^n = C_0 + C_1x + C_2x^2 + ........ | vedclass

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(B) We know that $(1 + x)^n = C_0 + C_1x + C_2x^2 + ..... + C_nx^n$ .....$(i)$Also,$(1 + \frac{1}{x})^n = C_0 + C_1\frac{1}{x} + C_2(\frac{1}{x})^2 +

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

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(B) We know that $(1 + x)^n = C_0 + C_1x + C_2x^2 + ..... + C_nx^n$ .....$(i)$Also,$(1 + \frac{1}{x})^n = C_0 + C_1\frac{1}{x} + C_2(\frac{1}{x})^2 + ..... + C_n(\frac{1}{x})^n$ ....$(ii)$Multiplying $(i)$ and $(ii)$,the sum $C_0^2 + C_1^2 + C_2^2 + ..... + C_n^2$ is the coefficient of the term independent of $x$ in the product $(1 + x)^n(1 + \frac{1}{x})^n$.This is equivalent to the coefficient of $x^n$ in the expansion of $\frac{1}{x^n}(1 + x)^{2n}$,which is the coefficient of $x^n$ in $(1 + x)^{2n}$.The coefficient of $x^n$ in $(1 + x)^{2n}$ is given by $^{2n}C_n = \frac{(2n)!}{n!n!}$.

If $(1 + x)^n = C_0 + C_1x + C_2x^2 + .......... + C_nx^n,$ then $C_0^2 + C_1^2 + C_2^2 + C_3^2 + ...... + C_n^2$ =

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