CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Find the sum of  $$1^{2}+(1^{2}+2^{2})+(1^{2}+2^{2}+3^{2})+.......$$


Solution

First term$$={1}^{2}$$

Second term$$={1}^{2}+{2}^{2}$$

Third term$$={1}^{2}+{2}^{2}+{3}^{2}$$...

$${n}^{th}$$ term$$={1}^{2}+{2}^{2}+{3}^{2}+...+{n}^{2}$$

Here $${a}_{n}={1}^{2}+{2}^{2}+{3}^{2}+...+{n}^{2}$$

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

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

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

Now, the sum of $$n$$ terms is 

$${S}_{n}=\sum_{n=1}^{n}{{a}_{n}}$$

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

$$=\dfrac{1}{6}\left(2\sum_{n=1}^{n}{{n}^{3}}+3\sum_{n=1}^{n}{{n}^{2}}+\sum_{n=1}^{n}{n}\right)$$

$$=\dfrac{1}{6}\left(\dfrac{2{n}^{2}{\left(n+1\right)}^{2}}{2}+\dfrac{3n\left(n+1\right)\left(2n+1\right)}{6}+\dfrac{n\left(n+1\right)}{2}\right)$$

$$=\dfrac{n\left(n+1\right)}{12}\left[n\left(n+1\right)+\left(2n+1\right)+1\right]$$

$$=\dfrac{n\left(n+1\right)}{12}\left[{n}^{2}+n+2n+1+1\right]$$

$$=\dfrac{n\left(n+1\right)}{12}\left[{n}^{2}+3n+2\right]$$

$$=\dfrac{n\left(n+1\right)}{12}\left[{n}^{2}+2n+n+2\right]$$

$$=\dfrac{n\left(n+1\right)}{12}\left[\left(n+1\right)\left(n+2\right)\right]$$

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

Thus, the required sum is $$\dfrac{n{\left(n+1\right)}^{2}\left(n+2\right)}{12}$$


Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More



footer-image