Find the sum `2xx5+5xx9+8xx13+11xx17+..` n terms.

Explanation

The first factors of terms are 2,5,8,11,…
This is an A.P. whose nth term is (3n-1).
The second factors of terms are 5,9,13,17,..
This is an A.P. whose nth term is (4n+1).
So, rth term of the given series,
`T_(r)=(3r-1)(4r+1)=12r^(2)-r-1`
Hence, sum of the series,
`S=underset(r=1)overset(n)sumT_(r)=sum_(r=1)^(n)(12r^(2)-r-1)`
`=sum_(r=1)^(n)12r^(2)-sum_(r=1)^(n)r-sum_(r=1)^(n)1`
`=12(n(n+1)(2n+1))/6-(n(n+1))/2-n`
`=n/2[8n^(2)+11n+1]`