A) 60
B) 120
C) 64
D) 124
Correct Answer: B
Solution :
[b] General term of the given series is \[r\frac{^{n}{{C}_{r}}}{^{n}{{C}_{r-1}}}=n+1-r\] By taking summation over n, we get \[\sum\limits_{1}^{15}{r\frac{^{n}{{C}_{r}}}{^{n}{{C}_{r-1}}}=\sum\limits_{n=1}^{15}{(n+1-r)=\sum\limits_{1}^{15}{(16-r)}}}\] \[=16\times 15-\frac{1}{2}\cdot 15\times 16\] By using sum of n natural numbers \[=\frac{n(n+1)}{2}\] \[=240-120=120\]You need to login to perform this action.
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