Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + …

Asked by Pragya Singh | 2 years ago |  119

##### Solution :-

The given series is 3 + 7 + 13 + 21 + 31 + …

S = 3 + 7 + 13 + 21 + 31 + …+ an–1 + an

S = 3 + 7 + 13 + 21 + …. + an – 2 + an – 1 + an

On subtracting both the equations, we get

S – S = [3 + (7 + 13 + 21 + 31 + …+ an–1 + an)] – [(3 + 7 + 13 + 21 + 31 + …+ an–1) + an]

S – S = 3 + [(7 – 3) + (13 – 7) + (21 – 13) + … + (an – an–1)] – an

0 = 3 + [4 + 6 + 8 + … (n –1) terms] – an

an = 3 + [4 + 6 + 8 + … (n –1) terms]

$$=3 +(\dfrac{n-1}{2})[2\times 4+(n-1-1)2]$$

$$3 +(\dfrac{n-1}{2})[8+(n-2)2]$$

$$3 +(\dfrac{n-1}{2})(2n+4)$$

$$=3+(n-1)(n+2)$$

$$=3+(n^2+n-2)$$

$$=n^2+n+1$$

Therefore,

$$\displaystyle\sum_{k=1}^{n} a_k=\displaystyle\sum_{k=1}^{n}k^2+\displaystyle\sum_{k=1}^{n} k+\displaystyle\sum_{k=1}^{n}1$$

$$\dfrac{n(n+1)(2n+1)}{6}+\dfrac{n(n+1)}{2}+n$$

$$n[ \dfrac{(n+1)(2n+1)+3(n+1)+6}{6}]$$

$$n[ \dfrac{2n^2+3n+13n+3+36}{6}]$$

$$n[ \dfrac{2n^2+6n+10}{6}]$$

$$\dfrac{n}{3}[n^2+3n+5]$$

Therefore, the sum of n terms of the given series

$$\dfrac{n}{3}[n^2+3n+5]$$

Answered by Abhisek | 2 years ago

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