Prove that $(C_{0} + C_{1} + C_{2} + . . . + C_{n})^{2} = 1 +^{2 n} C_{1} +^{2 n} C_{2} + . . . . +^{2 n} C_{n}$

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Solution

$(1 + x)^{n} + (1 + x)^{n} = (1 + x)^{2 n}$
Expand both the sides and put x = 1

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If $C_{0} + C_{1} + C_{2} + . . . + C_{n} = 256$ = then $^{2 n} C_{2}$ is equal to

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