Question

Find the sum to n terms of the series
1 square +1square+2square+1square+2square+3square

Answer 1

Answer:

20

Step-by-step explanation:

1square=1

2square=4

3square=9

1+1+4+1+4+9=20

Answer 2

The sum of the given series = $\dfrac{n(n+1)^2(n+2)}{12}$

Step-by-step explanation:

The given series:

$1^{2} +(1^{2}+2^{2})+(1^{2}+2^{2}+3^{2})+(1^{2}+2^{2}+3^2+4^2)+(1^{2}+2^{2}+3^2+4^2+...+n^2)$

To find, the sum of  n terms of the given series = ?

∴ $1^{2} +(1^{2}+2^{2})+(1^{2}+2^{2}+3^{2})+(1^{2}+2^{2}+3^2+4^2)+(1^{2}+2^{2}+3^2+4^2+...+n^2)$

We know that,

$\sum_{i=1}^{n}i=\dfrac{n(n+1)}{2}$,

$\sum_{i=1}^{n}i^2=\dfrac{n(n+1)(2n+1)}{6}$ and

$\sum_{i=1}^{n}i^3=(\dfrac{n(n+1)}{2})^2$

∴ $1^{2} +(1^{2}+2^{2})+(1^{2}+2^{2}+3^{2})+(1^{2}+2^{2}+3^2+4^2)+(1^{2}+2^{2}+3^2+4^2+...+n^2)$

$\sum_{i=1}^{n}(1^{2}+2^{2} +3^{2} +.......+i^{2})$

= $\sum_{i=1}^{n}\dfrac{i(i+1)(2i+1)}{6}$

$=\sum_{i=1}^{n}\dfrac{2i^3+3i^2+i}{6}$

$=\dfrac{1}{3} (\dfrac{n(n+1)}{2})^2+\dfrac{n(n+1)(2n+1)}{12}+\dfrac{n(n+1)}{12}$

= $\dfrac{n(n+1)^2(n+2)}{12}$

Thus, the sum of the given series = $\dfrac{n(n+1)^2(n+2)}{12}$