Find the sum to `n` terms of the series: `1+5+12+22+35+........`

Explanation

Let ` S_(n) = 1 + 5 + 12 + 22 + …+ a_(n)` (i)
Also , ` S_(n) = 1 + 5 + 12 + …+ a_(n-1) +a_(n)` (ii)
subracting (ii) from (i) , we get
` 0 = 1 + [ 4 + 7 + 10 + 13 + …+ (a_(n) - a_(n-a))] - a_(n) `
`rArr a_(n) = 1 + (n-1)/(2) [2xx4 + (n-1-1)xx3]`
` rArr a_(n) + ((n-1)/(2)) (3n+ 2)`
` therefore a_(n) = (1)/(2) (3n^(2) -n)`
` therefore a_(n) = (1)/(2) n (3n -1)`