Sum the series `3. 8+6. 11+9. 14+ ` to `n` terms.

Answer

N/a

Explanation

Given series `= 3 xx 8 + 6 xx 11 + 9 xx 14 + …+` to n terms
Then nth term of given series `T_(n)`
= (nth term of sequence 3, 6, 9, …) `xx ` (nth term of sequence 8, 11, 14, …)
`= [3 + (n-1)3] xx [8 + (n-1)3]`
`= 3n xx (3n + 5) = 9n^(2) + 15n`
Sum of n terms of the series
`S_(n) = sum(9n^(2) + 15n) = 9sumn^(2) + 15 sumn`
`= 9 (n(n+1)(2n+1))/(6) + 15(n(n + 1))/(2)`
`= (3n(n + 1)(2n + 1))/(2) + 5(3n(n + 1))/(2)`
`= (3n(n + 1))/(2)[2n + 1 + 5]`
`= (3n(n + 1))/(2) * 2(n + 3)`
`= 3n(n + 1)(n + 3)`
Therefore, sum of n terms of given series
`= 3n(n + 1)(n + 3)`