math

the first differences are : 4, 14, 30, 52, 80, 104...
2nd differences are: 10, 16, 22, 28, 34
3rd differences are 6,6,6,6,6,.. constant....

now bcos it's 3rd difference is const, the answer will be of the form?

polynomial. (which order? ) think... if u cant .. i will complete the solution...

otherwise just guess...

basically there is a rule..

i will explain..

in an AP, the first difference is a constant.
The sum is n/2(2a+nd-d)

this is 2nd degree polynomial in n

if the 2nd difference is const. the sum will be a 3rd degree poynomial.. and so on....

so our solution is of the form..
an⁴+bn³+cn²+dn+e

continued....

now we have many equations and 5 varaibles a,b,c,d,e

equation will be by puttin n=1 and sum = 1
n=2 , sum =1+5=6
n=3, sum=1+5+19=25
n=4, sum=1+5+19+49
...
...
.....

This I am sure u can solve by urself...

It requires no skills :)

or u could use matrices to solve this one!

I hope u got the whole solution...
the last part was very dirty.. getting 4th powers of 3,4 5, etc.. so i did not want to dirty my hands...

IIT JEE will never ask such problems.... (I mean with such calculations)

So the important take away from this problem is that u need to understand how to reach that point...