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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy
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Application +4 | -1
If the function f:R→Rf: R \rightarrow R satisfies f(x+y)=f(x)+f(y),∀ x,y∈Rf(x+y)=f(x)+f(y), \forall \, x, y \in \mathrm{R} and f(1)=7f(1)=7, then ∑r=1nf(r)\sum\limits_{{r=1}}^n f(r)
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Correct option: D
7n(n+1)2\frac{7 n(n+1)}{2}
Option D (7n(n+1)2\frac{{7 n(n+1)}}{{2}}) is correct because the function f(r) equals 7r, and the sum of 7r from r=1 to n is 7n(n+1)2\frac{{7 n(n+1)}}{{2}}.

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