find-the-value-of-p-pls-help-1-3-3-3-5-3-p-3-8128-

Question Number 104706 by Skabetix last updated on 23/Jul/20

f i n d t h e v a l u e o f p (p l s h e l p)

1^{3} + 3^{3} + 5^{3} + \dots + p^{3} = 8128

Commented by mr W last updated on 23/Jul/20

Commented by mr W last updated on 23/Jul/20

n^{2} (2 n^{2} - 1) = 8128

2 n^{4} - n^{2} - 8128 = 0

n^{2} = \frac{1 + \sqrt{1 + 4 \times 2 \times 8128}}{4} = 64

n = 8

p = 2 n - 1 = 2 \times 8 - 1 = 15

Answered by Dwaipayan Shikari last updated on 23/Jul/20

T_{n} = {(2 n - 1)}^{3} = 8 n^{3} - 1 - 12 n^{2} + 6 n

Σ T_{n} = 2 {(n (n + 1))}^{2} - 2 n (n + 1) (2 n + 1) + 3 n (n + 1) - n

= 2 n (n + 1) (n^{2} + n - 2 n - 1) + n (3 n + 2)

= n ((2 n + 2) (n^{2} - n - 1)) + n (3 n + 2)

= n (2 n^{3} + 2 n^{2} - 2 n^{2} - 2 n - 2 n - 2 + 3 n + 2)

= n (2 n^{3} - n) = n^{2} (2 n^{2} - 1)

n^{2} (2 n^{2} - 1) = 8128

\Rightarrow a (2 a - 1) = 8128

\Rightarrow 2 a^{2} - a - 8128 = 0

a = 64

n = 8

T_{n} = (2 n - 1) = 15