A) \[^{n+1}{{C}_{k+1}}\]
B) \[^{n}{{C}_{k}}\]
C) \[^{n}{{C}_{n-k-1}}\]
D) None of these
Correct Answer: A
Solution :
The expression being in G. P. \[E=1+(1+x)+{{(1+x)}^{2}}+....+{{(1+x)}^{n}}\] \[\frac{{{(1+x)}^{n+1}}-1}{(1+x)-1}={{x}^{-1}}\{{{(1+x)}^{n+1}}-1\}\] \[\therefore \,\,\,\]The coefficient of x k in E = The coefficient of \[{{x}^{k+1}}\]in\[\{{{(1+x)}^{n+1}}-1\}\] \[={{\,}^{n+1}}{{C}_{k+1}}\].
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