Question

The sum of first 9 terms of the series $\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+.........$
(a) 142
(b) 192
(c) 71
(d) 96

Answer 1

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Answer 2

Numerator :

The sum of cubes of natural numbers,

Hence, n-th term has the Numerator

Σn³

Denominator :

The sum of odd natural numbers,

Hence n-th term has the Denominator

Σ (2n-1)

nth term = Σ n³ / Σ (2n+1)

⇒sum of cubes of n natural numbers Σ n³ = n² ( n + 1)² / 4

⇒sum of n odd natural numbers Σ 2(n)-1 = n²

So n th term is

$= \dfrac{ \dfrac{ {n}^{2} (n + 1) ^{2} }{4} }{ {n}^{2} } \\ \\ = \frac{(n + 1) ^{2} }{4}$

Now We need to find the sum till 9 terms.

$= \sum \frac{(n + 1) ^{2} }{4} \\ \\ = \sum \frac{1}{4} ( {n}^{2} + 1 + 2n) \\ \\ = \frac{1}{4} ( \sum {n}^{2} + \sum \: 1 + 2\sum \: n) \\ \\ = \frac{1}{4} (9 + \frac{9 \times 10 \times 19}{6} + 9 \times 10) \\ \\ = \frac{1}{4} (9 + 90 + 285) \\ \\ = \frac{1}{4} (99 + 285) \\ \\ = \frac{384}{4} \\ \\ = 96$

Therefore, The sum to 9 terms is 96.