If $S_{n} = n^{2} + 2 n,$ find $T_{n}$

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Solution

Given $S_{n} = n^{2} + 2 n = n (n + 2)$
$S_{n - 1} = (n - 1) (n - 1 + 2) = (n - 1) (n + 1) = n^{2} - 1$
$S_{n} = S_{n - 1} + T_{n}$
$⟹ T_{n} = S_{n} - S_{n - 1} = n^{2} + 2 n - (n^{2} - 1) = 2 n + 1$

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Similar questions

If $S_{n} = \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . . . + \frac{1}{n^{2}}$ and $T_{n} = \frac{2 n - 1}{n}$ , then

S_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}}

t_{n} = \sum_{r = 0}^{n} \frac{r}{^{n} C_{r}},

\frac{t_{n}}{S_{n}}

t_{n} = n^{2} - 2 n

\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}}

\frac{t_{m}}{t_{n}} =

S_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}}

t_{n} = \sum_{r = 0}^{n} \frac{r}{^{n} C_{r}},

\frac{t_{n}}{S_{n}}

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