Cubes inside roots?

Calculus Level 2

1 1 3 + 1 1 3 + 2 3 + 1 1 3 + 2 3 + 3 3 + . . . + 1 1 3 + 2 3 + . . . + 9 9 3 \sqrt{\frac{1}{1^3}}+\sqrt{\frac{1}{1^3+2^3}}+\sqrt{\frac{1}{1^3+2^3+3^3}}+...+\sqrt{\frac{1}{1^3+2^3+...+99^3}}

What is the value of the expression above?


The answer is 1.98.

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Ivander Jonathan by Ivander Jonathan , 21, Indonesia

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2 solutions

Discussions for this problem are now closed

Noting that ∑ i = 1 n i 3 = ( n ( n + 1 ) 2 ) 2 \sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2 ,

the required sum would be

∑ k = 1 99 2 k ( k + 1 ) = 2 ∑ k = 1 99 ( 1 k − 1 k + 1 ) \sum_{k=1}^{99} \frac{2}{k(k+1)} = 2\sum_{k=1}^{99} \left(\frac{1}{k}-\frac{1}{k+1}\right)

= 2 ∑ k = 1 99 1 k − 2 ∑ k = 2 100 1 k =2\sum_{k=1}^{99} \frac{1}{k} - 2\sum_{k=2}^{100} \frac{1}{k}

= 2 ( 1 1 − 1 100 ) = 1.98 = 2\left(\frac{1}{1}-\frac{1}{100}\right)=\boxed{1.98}

34 Helpful 0 Interesting 0 Brilliant 0 Confused

I can't understand how you went from the first part to the required sum. Could you explain to me?

Spikin Parker - 6 years, 4 months ago

The first part simply states that the sum of the first n n cubes is equal to the n th n^{\text{th}} triangular number squared. So, for the n th n^{\text{th}} term of the required sum, we have ( 1 ( n ) ( n + 1 ) 2 ) 2 = 2 ( n ) ( n + 1 ) \left (\sqrt{\frac{1}{\frac{(n)(n+1)}{2}}}\right )^2 = \frac{2}{(n)(n+1)} . Then Janardhanan gets into some simple telescoping business. Hope this helped!

Ryan Tamburrino - 6 years, 4 months ago

After that????? Please explain....

BISWAJIT MOHANTY - 6 years, 2 months ago

Read more about finding the sum of telescoping series.

Justin Augustine - 6 years, 4 months ago
Omkar Kulkarni
Feb 1, 2015

I started off with ∑ n = 1 99 ( 1 ∑ k = 1 n k 3 ) \displaystyle \sum_{n=1}^{99} \left(\sqrt{\frac{1}{\displaystyle \sum_{k=1}^{n} k^{3}}}\right) from where onwards I simplified a little and then onto the same tracks of those as the above solution.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Although you're on the right track, please refrain from a posting a solution when the working is inadequate. You didn't mention a closed form for ∑ k = 1 99 k 3 \sum_{k=1}^{99} k^3 is attainable and you should at least give a hint of telescoping series in your answer.

I'm sorry, but I don't think so. I don't know the answer yet, but this process is wrong. Why? Well, think in we'll get differents results in each operation. I mean: sqrt(1/1^3) it's not the same that sqrt[1/(1^3 + 2^3)] Your answer suggests that we'll get always (99 times) the last result: sqrt[1/(1^3 + 2^3 +...+ 99^3)].

Spikin Parker - 6 years, 4 months ago

No, not 99 99 times the last result. The notation represents the given sum. Notice that the summation in the denominator is ∑ k = 1 n \displaystyle \sum_{k=1}^{\boxed{n}} which means that you start adding the cubes of the numbers till n n , and then a new square root begins all over from 1 1 .

Read more about telescoping at here , and then try this for practice.

Omkar Kulkarni - 6 years, 4 months ago

I can see better your idea now. Thanks.

Spikin Parker - 6 years, 4 months ago

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