Find the sum of the following series up to n terms:

$$\dfrac{1^3}{1}+ \dfrac{1^3+2^3}{1+3}+ \dfrac{1^3+2^3+3^3}{1+3+5}+....$$

Asked by Pragya Singh | 1 year ago |  64

##### Solution :-

$$\dfrac{1^3}{1}+ \dfrac{1^3+2^3}{1+3}+ \dfrac{1^3+2^3+3^3}{1+3+5}+....$$

$$\dfrac{[\dfrac{n(n+1)}{2}]^2}{1+3+5+...+(2n-1)}$$ is the nth term of the given series.

With first term a , last term (2n -1) and number of terms as n then 1,3,5...(2n -1) is an A.P.

Therefore,

$$1+3+5+...+(2n-1)$$

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

Also,

$$\dfrac{n^2(n+1)^2}{4n^2}=\dfrac{(n+1)^2}{4}$$

$$\dfrac{1}{4}n^2+\dfrac{1}{4}n+\dfrac{1}{4}$$

And,

$$\displaystyle\sum_{k=1}^{n}a_k=\displaystyle\sum_{k=1}^{n} (\dfrac{1}{4}k^2+\dfrac{1}{2}k+\dfrac{1}{4})$$

$$\dfrac{1}{4}\dfrac{n(n+1)(2n+1)}{6}+\dfrac{1}{2}\dfrac{n(n+1)}{2}+\dfrac{1}{4}n$$

$$\dfrac{n[(n+1)(2n+1)+6(n+1)+6]}{24}$$

$$\dfrac{n[2n^2+3n+1+6n+6+6]}{24}$$

=$$\dfrac{n[2n^2+9n+13]}{24}$$

Therefore, the sum of n terms of the given series

$$\dfrac{n[2n^2+9n+13]}{24}$$

Answered by Abhisek | 1 year ago

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