Find the sum `3+7+14+24+37+…..20 ` terms

Answer

4240

Explanation

Clearly, here the difference between the successive terms are
7-3,14-7,24-14,..
i.e., 4,7,10,… which are in A.P.
`thereforeT_(n)=an^(2)+bn+c`
Thus, we have
3=a+b+c,
7=4a+2b+c
and 14=9a+3b+c
Solving we get,
`a=3/2,b=-1/2,c=2`
Hence, `T_(n)=1/2(3n^(2)-n+4)`
`thereforeS_(n)=1/2[3Sigman^(2)-Sigman+4n]`
`=1/2[3(n(n+1)(2n+1))/6-(n(n+1))/2+4n]`
`=n/2(n^(2)+n+4)`
`rArrS_(20)=4240`