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Question: If \(f:R \rightarrow R\) satisfies \(f(x + y) = f(x) + f(y),\) for all \(x,y \in R\) and \(f(1) = 7,...

If f:R→Rf:R \rightarrow R satisfies f(x+y)=f(x)+f(y),f(x + y) = f(x) + f(y), for all x,y∈Rx,y \in R and f(1)=7,f(1) = 7, then ∑r=1nf(r)\sum_{r = 1}^{n}{f(r)} is

A

7n2\frac{7n}{2}

B

7(n+1)2\frac{7(n + 1)}{2}

C

7n(n+1)7n(n + 1)

D

7n(n+1)2\frac{7n(n + 1)}{2}

Answer

7n(n+1)2\frac{7n(n + 1)}{2}

Explanation

Solution

f(x+y)=f(x)+f(y)f(x + y) = f(x) + f(y)

put x=1,y=0x = 1,y = 0 ⇒ f(1)=f(1)+f(0)=7f(1) = f(1) + f(0) = 7

put x=1,y=1x = 1,y = 1 ⇒ f(2)=2.f(1)=2.7f(2) = 2.f(1) = 2.7; similarly f(3)=3.7f(3) = 3.7 and so on

∴∑r=1nf(r)=7(1+2+3+.....+n)\therefore\sum_{r = 1}^{n}{f(r) = 7(1 + 2 + 3 + ..... + n)} = 7n(n+1)2\frac{7n(n + 1)}{2}.