Find the sum of 40 terms of the series 1 + 5 + 12 + 22 + 35 + .................

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Solution

Sequence of difference between two consecutive terms is 4,7,10,18............This is an AP.

Given $S_{n}$ = 1 + 5 + 12 + 22 + 35 --------------------- $t_{n}$ .......................(1)

Write the $S_{n}$ in such a way that $1^{s t}$ term of equation 2 comes under $2^{n d}$ term of equation 1

$S_{n}$ = 1 + 5 + 12 + 22 --------------------- $t_{n - 1}$ + $t_{n}$ .......................(2)

Subtracting equation (2) from equation (1)

0 = 1 + 4 + 7 + 10 + 13 .................( $T_{n} - T_{n - 1}$ ) - $T_{n}$

$T_{n}$ = 1 + 4 + 7 + 10 + 13.................n terms

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

= $\frac{n}{2}$ [2 + (n-1)3] = $\frac{3}{2}$ $n^{2}$ - $\frac{n}{2}$

Sum of n terms $S_{n} = \sum_{i = 1}^{n} t_{n}$

= $\sum_{i = 1}^{n}$ $(\frac{3}{2} n^{2} - \frac{n}{2})$

= $\frac{3}{2}$ $\sum_{i = 1}^{n}$ $n^{2}$ - $\frac{1}{2}$ $\sum_{i = 1}^{n}$ n

substitute n=40, $S_{40} = 32800$

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Find the sum of 40 terms of the series 1 + 5 + 12 + 22 + 35 + ................. __

Find the sum of 40 terms of the series 1 + 5 + 12 + 22 + 35 + .................

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