Given, f(x+y)=f(x)+f(y)+xy∀x,y∈R
Put y=1, we get
⇒f(x+1)=f(x)+f(1)+x
⇒f(x+1)−f(x)=3+x
On taking Sigma on both sides
⇒x=1∑n−1(f(x+1)−f(x))=x=1∑n−1(3+x)
⇒(f(2)−f(1))+(f(3)−f(2))+(f(4)−f(3))+(f(5)−f(4))+......+(f(n)−f(n−1))
=3(n−1)+2(n−1)n
⇒f(n)−f(1)=3(n−1)+2(n−1)n
⇒f(n)−3=3(n−1)+2(n−1)n
⇒f(n)=2n2+25n
Again taking Sigma on both the sides
⇒n=1∑10f(n)=n=1∑10(2n2+25n)
⇒n=1∑10f(n)=21.610.11.21+25.210.11
⇒n=1∑10f(n)=330.