A) 8
B) 4
C) 6
D) 5
Correct Answer: D
Solution :
[d] \[\sum\limits_{r=0}^{n}{\frac{r+2}{r+1}{{\,}^{n}}{{C}_{r}}=\sum\limits_{r=0}^{n}{\frac{r+1+1}{r+1}{{\,}^{n}}{{C}_{r}}}}\] |
\[=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}+\sum\limits_{r=0}^{n}{\frac{{{\,}^{n}}{{C}_{r}}}{r+1}={{2}^{n}}+\sum\limits_{r=0}^{n}{\frac{^{n+1}{{C}_{r+1}}}{n+1}}}}\] |
\[={{2}^{n}}+\frac{1}{n+1}\sum\limits_{r+0}^{n}{^{n+1}{{C}_{r+1}}}\] |
\[={{2}^{n}}+\frac{1}{n+1}({{2}^{n+1}}-1)\] |
\[=\frac{1}{n+1}[(n+1){{2}^{n}}+{{2}^{n+1}}-1]\] |
\[=\frac{1}{n+1}[{{2}^{n}}(n+3)-1]\] |
Given, \[\frac{(n+3){{2}^{n}}-1}{n+1}=\frac{{{2}^{8}}-1}{6}=\frac{(5+3){{.2}^{5}}-1}{5+1}\] |
\[\Rightarrow \,\,\,\,n=5\] |
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