Sum of the given series will be

Calculus Level 3

7 2 3 + 19 6 3 + 37 1 2 3 + 61 2 0 3 + ⋯ = ? \large \frac 7{2^3} + \frac {19}{6^3} + \frac {37}{12^3} + \frac {61}{20^3} + \cdots = \ ?

Note that

  • In the sequence 7 , 19 , 37 , 61 , … 7,19,37,61,\ldots , the difference between each consecutive terms follows an arithmetic progression.

  • In the sequence 2 , 6 , 12 , 20 , … 2,6,12,20, \ldots ,the difference between each consecutive terms follows an arithmetic progression as well.

1 2 3 4

Gaurav Raj by Gaurav Raj , 28, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

S = 7 2 3 + 19 6 3 + 37 1 2 3 + 61 2 0 3 + ⋯ = 1 + 6 ( 1 ⋅ 2 ) 3 + 1 + 6 + 2 ⋅ 6 ( 2 ⋅ 3 ) 3 + 1 + 6 + 2 ⋅ 6 + 3 ⋅ 6 ( 3 ⋅ 4 ) 3 + 1 + 6 + 2 ⋅ 6 + 3 ⋅ 6 + 4 ⋅ 6 ( 4 ⋅ 5 ) 3 + ⋯ = lim ⁡ n → ∞ ∑ k = 1 n 1 + 6 ⋅ k ( k + 1 ) 2 ( k ( k + 1 ) ) 3 = lim ⁡ n → ∞ ∑ k = 1 n 3 k 2 + 3 k + 1 k 3 ( k + 1 ) 3 = lim ⁡ n → ∞ ∑ k = 1 n k 3 + 3 k 2 + 3 k + 1 − k 3 k 3 ( k + 1 ) 3 = lim ⁡ n → ∞ ∑ k = 1 n ( k + 1 ) 3 − k 3 k 3 ( k + 1 ) 3 = lim ⁡ n → ∞ ∑ k = 1 n ( 1 k 3 − 1 ( k + 1 ) 3 ) = lim ⁡ n → ∞ ( 1 − 1 ( n + 1 ) 3 ) = 1 \begin{aligned} S & = \frac 7{2^3} + \frac {19}{6^3} + \frac {37}{12^3} + \frac {61}{20^3} + \cdots \\ & = \frac {1+6}{(1\cdot2)^3} + \frac {1+6+2\cdot 6}{(2\cdot 3)^3} + \frac {1+6+2\cdot 6+3\cdot 6}{(3\cdot 4)^3} + \frac {1+6+2\cdot 6+3\cdot 6+4\cdot 6}{(4\cdot 5)^3} + \cdots \\ & = \lim_{n \to \infty} \sum_{k=1}^n \frac {1+6\cdot\frac {k(k+1)}2}{(k(k+1))^3} = \lim_{n \to \infty} \sum_{k=1}^n \frac {3k^2+3k+1}{k^3(k+1)^3} = \lim_{n \to \infty} \sum_{k=1}^n \frac {k^3+3k^2+3k+1 - k^3}{k^3(k+1)^3} \\ & = \lim_{n \to \infty} \sum_{k=1}^n \frac {(k+1)^3 - k^3}{k^3(k+1)^3} = \lim_{n \to \infty} \sum_{k=1}^n \left(\frac 1{k^3} - \frac 1{(k+1)^3} \right) = \lim_{n \to \infty} \left(1 - \frac 1{(n+1)^3} \right) = \boxed{1} \end{aligned}

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Gaurav Raj , use ⋯ \cdots will do. It indicates that there are infinite terms. The ∞ \infty at the end is unnecessary. It can mean that the last term is ∞ \infty instead of 0 as in this case. Use only the standard three dots ⋯ \cdots will do not 5, 6, 19 and all that.

Chew-Seong Cheong - 3 years, 3 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...