A) \[\frac{7n}{2}\]
B) \[\frac{7(n+1)}{2}\]
C) \[7n\,(n+1)\]
D) \[\frac{7n\,(n+1)}{2}\]
Correct Answer: D
Solution :
\[\sum\limits_{r=1}^{n}{f(r)=f(1)+f(2)+f(3)+...+f(n)}\] \[=f(1)+2f(1)+3f(1)+....+nf(1)\] [since, \[f(x+y)=f(x)+f(y)\]] \[=(1+2+3+...+n)\,f(1)\] \[\frac{=n(n+1)}{2}.7=\frac{7n(n+1)}{2}\]You need to login to perform this action.
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