# Find the sum of n terms of the series 3 + 7 + 14 + 24 + 37 + …….

## Solution :

By using method of differences,

The difference between the successive terms are 7 – 3 = 4, 14 – 7 = 7, 24 – 14 = 10, …. Clearly,these differences are in AP.

Let $$T_n$$ be the nth term and $$S_n$$ denote the sum to n terms of the given series

Then, $$S_n$$ = 3 + 7 + 14 + 24 + 37 + ….. + $$T_{n-1}$$ + $$T_n$$ ……(i)

Also, $$S_n$$ =         3 + 7 + 14 + 24 + 37 + ….. + $$T_{n-1}$$ + $$T_n$$ ……(ii)

Subtracting (ii) from (i), we get

0 = 3 + [4 + 7 + 10 + 13 + …… + $$T_{n-1}$$ + $$T_n$$] – $$T_n$$

$$\implies$$ $$T_n$$ = 3 + $$(n-1)\over 2$${2*4+(n-1-1)*3} = $$6 + (n-1)(3n+2)\over 4$$

$$\implies$$ $$T_n$$ = $$1\over 2$$ ($$3n^2 – n + 4$$)

$$\therefore$$ $$S_n$$ = $$\sum_{k=1}^{n}$$ $$T_k$$ = $$\sum_{k=1}^{n}$$ $$1\over 2$$ ($$3n^2 – n + 4$$) = $$1\over 2$$[$$\sum_{k=1}^{n}$$$$3k^2$$ – $$\sum_{k=1}^{n}$$ k + 4n]

$$\implies$$ $$S_n$$ = $$1\over 2$$[3$$n(n+1)(2n+1)\over 6$$  – $$n(n+1)\over 2$$ + 4n] = $$n\over 2$$ ($$n^2 + n + 4$$)

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