Find sum: $1 + 5 + 12 + 22 + 35 + . . . . .$

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Solution

Let
$S_{n} = 1 + 5 + 12 + 22 + 35 + . . . . . . . . . S_{n} = 1 + 5 + 12 + 22 + . . . . . . - - - - -____________________________$
$\Rightarrow t_{n} = 1 + 4 + 7 + 10 + 13 + . . . . . . . n$ terms
$\Rightarrow t_{n} = \frac{n}{2} {2 \times 1 + (n - 1) 3}$
$= \frac{n}{2} (2 + 3 n - 3)$
$= \frac{n}{2} (3 n - 1)$
$t_{n} = \frac{3 n^{2}}{2} - \frac{3 n^{2}}{2} - \frac{n}{2}$
$S_{n} = n \sum n = 1 t_{n}$
$= \frac{3}{2} \times \frac{n (n + 1) (2 n + 1)}{6} \frac{- 1}{2} \frac{n (n + 1)}{2}$
$= \frac{n (n + 1)}{4} [\frac{3 (2 n + 1)}{3} - 1]$
$= \frac{n (n + 1)}{4} \times 2 n$
$∴ S_{n} = \frac{n^{2} (n + 1)}{2}$
Hence, $S_{n} = \frac{n^{2} (n + 1)}{2}$

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Find sum: 1+5+12+22+35+.....

Let Sn=1+5+12+22+35+.........Sn= 1+5+12+22+......−−−−−____________________________⇒tn=1+4+7+10+13+.......n terms⇒tn=n2{2×1+(n−1)3}=n2(2+3n−3)=n2(3n−1)tn=3n22−3n22−n2Sn=n∑n=1tn=32×n(n+1)(2n+1)6−12n(n+1)2=n(n+1)4[3(2n+1)3−1]=n(n+1)4×2n∴Sn=n2(n+1)2Hence, Sn=n2(n+1)2

Find sum: $1 + 5 + 12 + 22 + 35 + . . . . .$