How to do this sum regarding series?

Question

Can anyone explain how I can do this sum?

The sum of first $n $ terms of a series is $n^2 +an+b $. Show that $b=0$ and the series is in A.P.

Answer 1

Let $T(n)$ be the $n$th term.

\begin{align*} T(n)&=[T(1)+T(2)+\cdots+T(n)]-[T(1)+T(2)+\cdots+T(n-1)]\\ &=(n^2+an+b)-[(n-1)^2+a(n-1)+b]\\ &=2n-1+a \end{align*}

$T(n)-T(n-1)=2$. So it is an A.P. and

$$T(1)+T(2)+\cdots+T(n)=\frac{n}{2}[(1+a)+(2n-1+a)]$$

which is divisible by $n$. So $b=0$.