A) 225
B) 330
C) 165
D) 190
Correct Answer: B
Solution :
\[f(x)=a{{x}^{2}}+bx+c\] \[f(1)=a+b+c=3\] Now \[f(x+y)=f(x)+f(y)+xy\] put \[y=1\] \[f(x+1)=f(x)+f(1)+x\] \[f(x+1)=f(x)+x+3\] Now \[f(2)=7\] \[f(3)=12\] Now \[{{S}_{n}}=3+7+12+......{{t}_{n}}\] ?.. \[{{S}_{n}}=3+7+......{{t}_{n-1}}+{{t}_{n}}\] ?.. On subtracting from \[{{t}_{n}}=3+4+5+.....\]upto n terms \[{{t}_{n}}=\frac{({{n}^{2}}+5n)}{2}\] \[{{S}_{n}}=\sum{{{t}_{n}}}=\sum{\frac{({{n}^{2}}+5n)}{2}}\] \[{{S}_{n}}=\frac{1}{2}\left[ \frac{n(n+1)(2n+1)}{6}+\frac{5n(n+1)}{2} \right]\]\[{{S}_{10}}=330\]
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