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Find sum of 10 terms of series 2+5+11+20+32+.....


 

Option: 1

151


Option: 2

515


Option: 3

505


Option: 4

490


Answers (1)

Difference of consecutive terms are

3,6,9,12+..... which is an AP.

Let S_n = 2+5+11+20+.......+t_{n-1}+t_n

and S_n = 2+5+11+.......+t_{n-2}+t_{n-1}+t_n

Subtract:

0= 2+(3+6+9+........)-t_n

                                        (n-1) terms

\Rightarrow t_n= 2+(3+6+9+........)

\Rightarrow t_n= 2+\frac{n-1}{2}[6+(n-2)3]

\Rightarrow t_n= 2+\frac{3(n-1)(n)}{2}

\Rightarrow t_n=2+\frac{3}{2}n^2+\frac{3}{2}n

\Rightarrow \sum t_n=2\sum 1+\frac{3}{2}\sum n^2+\frac{3}{2}\sum n

\Rightarrow \sum t_n=2n+\frac{3}{2}.\frac{n(n+1)(2n+1)}{6}-\frac{3}{2}.\frac{n(n+1)}{2}

So, S_{10}=20+\frac{3}{2}.\frac{10.11.21}{6}-\frac{3}{2}.\frac{10.11}{2}

               =515

Posted by

Sumit Saini

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