Question

How many terms of the series 1³ + 2³ + 3³ + … should be taken to get the sum 14400?

Answer 1

1³ + 2³ + 3³ + ... + n³ = 14400

`(("n"("n" + 1))/2)^2` = 14400 = (120)² 

`("n"("n" + 1))/2 = sqrt(14400)` = 120

n(n + 1) = 240

Method 1:

n² + n – 240 = 0

n² + 16n – 15n – 240 = 0

n(n + 16) – 15(n + 16) = 0

(n + 16)(n – 15) = 0

n = – 16, 15

∴ 15 terms to be taken to get the sum 14400.

Method 2:

n² + n – 240 = 0

n = `(- "b" ± sqrt("b"^2 - 4"ac"))/(2"a")`

= `(-1 ± sqrt(1 - 4 xx 1 xx - 240))/(2 xx 1)`

= `(- 1 ± sqrt(1 + 960))/2`

= `(- 1 ± sqrt(961))/2`

= `(- 1 ± 31)/2`

= `(-1 + 31)/2` or `(-1 - 31)/2`

= `30/2` or `(-32)/2`

n = 15 or – 16

n cannot be – 16

∴ n = 15