This is the question which I am referring to
If $S_n=an^2 + bn $ , verify that the series $\sum {t_{n}}$ is arithmetic where $ S_n=\sum_{n=1}^{n}{t_n} $
My try:
- first of all I used below equation to calculate ${t_n}$
${t_n} = S_n - S_{n-1} = a (2n-1)-b$
Then I calculated common difference by below equation
$d=t_n-t_{n-1}$
- Then I calculated $t_1$ by putting $n=1$.
- Then I calculated $s_n$ by AP formula and it came out to be
$s_n= an^2-bn$ but for $\sum {t_{n}}$ to be arithmetic $S_n=s_n$
Please mention where am I wrong
Answer
$t_n=a\{n^2-(n-1)^2\}+b\{n-(n-1)\}=2na-a+b$
$$t_n-t_{n-1}=2na-a+b-\{2(n-1)a-a+b\}=2a$$ which being independent of $n$ is constant
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