Question

The sum of the series 3.6 + 4.7 + 5.8 + ....... upto (n – 2) terms

Options

• n³ + n² + n + 2

• `1/6 (2n^3 + 12n^2 + 10n - 84)`

• n³ + n² + n

• None of these

Answer 1

`1/6 (2n^3 + 12n^2 + 10n - 84)`

Explanation:

Given series 3.6 + 4.7 + 5.8 + ...... upto (n – 2) terms

T_r = (r + 2)(r + 5)

= r² + 7r + 10

S_n = `sumT_r`

= `sum_(r = 1)^(n - 2) (r^2 + 7r + 10)`

= `((n - 2)(n - 1)(2n - 3))/6 + (7(n - 2)(n - 1))/2 + 10(n - 2)`

= `([(n - 2)(n - 1)(2n - 3) + 21(n - 1)(n - 2) + 60(n - 2)])/6`

= `1/6 (n - 2)[(n - 1)(2n - 3) + 21(n - 1) + 60]`

= `1/6 (n - 2)[2n^2 - 2n - 3n + 3 + 21n - 21 + 60]`

= `1/6 (n - 2) [ 2n^2 + 16n + 42]`

= `1/6 (2n^3 + 16n^2 + 42n - 4n^2 - 32n - 84)`

= `1/6 (2n^3 +12n^2 + 10n - 84)`