A) \[\frac{{{3}^{11}}-1}{11}\]
B) \[\frac{{{2}^{11}}-1}{11}\]
C) \[\frac{{{11}^{3}}-1}{11}\]
D) \[\frac{{{11}^{2}}-1}{11}\]
Correct Answer: A
Solution :
We have, \[2.{{C}_{0}}+\frac{{{2}^{2}}}{2}.{{C}_{1}}+\frac{{{2}^{3}}}{2}{{C}_{2}}+....+\frac{{{2}^{11}}}{11}.{{C}_{10}}\] \[\sum\limits_{r=0}^{10}{^{10}{{C}_{r}}}.\frac{{{2}^{r+1}}}{r+1}=\frac{1}{11}\sum\limits_{r=0}^{10}{\frac{11}{r+1}}.\] \[^{10}C\] \[=\frac{1}{11}\sum\limits_{r=0}^{10}{^{11}{{C}_{r+1}}{{.2}^{r+1}}}\] \[=\frac{1}{11}{{(}^{11}}{{C}_{1}}{{.2}^{1}}+...{{+}^{11}}{{C}_{11}}{{.2}^{11}})\] \[=\frac{1}{11}{{(}^{11}}{{C}_{0}}{{.2}^{0}}{{+}^{11}}{{C}_{1}}{{.2}^{1}}\] \[+....{{+}^{11}}{{C}_{11}}{{.2}^{11}}{{-}^{11}}{{C}_{0}}{{.2}^{0}})\] \[=\frac{1}{11}[{{(1+2)}^{11}}-1]=\frac{{{3}^{11}}-1}{11}\]
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