Odd Cubes

Algebra Level 2

1 3 + 3 3 + 5 3 + 7 3 + ⋯ + ( 2 n − 1 ) 3 = ? \large 1^3 + 3^3 + 5^3 + 7^3 + \cdots + (2n-1)^3=\, ?

( ( n − 1 ) n 2 ) 2 \left( \frac{(n-1)n } { 2} \right)^2 ( ( 2 n + 1 ) n ) 2 \left( (2n+1)n \right)^2 2 n 2 ( n 2 − 1 ) 2n^2 (n^2 -1 ) n 2 ( 2 n 2 − 1 ) n^2 ( 2n^2 - 1 )

View wiki
Servia Rufus by Servia Rufus , 19, USA

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.

3 solutions

Chew-Seong Cheong
Nov 30, 2015

Let the given expression be S S , then we have:

S = 1 3 + 3 3 + 5 3 + 7 3 + . . . + ( 2 n − 1 ) 3 = 1 3 + 2 3 + 3 3 + 4 3 + . . . + ( 2 n − 1 ) 3 − 2 3 − 4 3 − 6 3 − 8 3 − . . . − ( 2 n − 2 ) 3 = 1 3 + 2 3 + 3 3 + 4 3 + . . . + ( 2 n − 1 ) 3 − 2 3 ( 1 3 + 2 3 + 3 3 + 4 3 + . . . + ( n − 1 ) 3 ) = ∑ k = 1 2 n − 1 k 3 − 8 ∑ k = 1 n − 1 k 3 = ( ( 2 n − 1 ) ( 2 n ) 2 ) 2 − 8 ( ( n − 1 ) ( n ) 2 ) 2 = n 2 ( 2 n − 1 ) 2 − 2 n 2 ( n − 1 ) 2 = n 2 ( 4 n 2 − 4 n + 1 ) − 2 n 2 ( n 2 − 2 n + 1 ) = n 2 ( 2 n 2 − 1 ) \begin{aligned} S & = 1^3 + 3^3 + 5^3 +7^3 +...+(2n-1)^3 \\ & = 1^3 + 2^3 + 3^3 +4^3 +...+(2n-1)^3 - 2^3 - 4^3 - 6^3 - 8^3 -...- (2n-2)^3 \\ & = 1^3 + 2^3 + 3^3 +4^3 +...+(2n-1)^3 - 2^3(1^3 + 2^3 + 3^3 +4^3 +...+(n-1)^3) \\ & = \sum_{k=1}^{2n-1} k^3 - 8 \sum_{k=1}^{n-1} k^3 \\ & = \left( \frac{(2n-1)(2n)}{2} \right)^2 - 8 \left( \frac{(n-1)(n)}{2} \right)^2 \\ & = n^2(2n-1)^2 - 2n^2(n-1)^2 \\ & = n^2(4n^2-4n+1) - 2n^2(n^2-2n+1) \\ & = \boxed{n^2(2n^2-1)} \end{aligned}

12 Helpful 0 Interesting 2 Brilliant 0 Confused

Yeah same way ... Nice solution..

A Former Brilliant Member - 5 years, 6 months ago
Deepak Kumar
Nov 30, 2015

Since options are given,just put n=1

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Yeah this approach is good from examination point of view but for improving the skills we need to get exact approach . but anyways nice solution

Prakhar Bindal - 5 years, 6 months ago

N=1 will eliminate 4 out of the 5 possible answers. Even knowing how to solve the question "properly", this is the easiest and thus correct way.

Jason Short - 5 years, 6 months ago
Akshat Sharda
Nov 30, 2015

∑ m = 1 n ( 2 m − 1 ) 3 ∑ m = 1 n ( 8 m 3 − 1 − 3 ⋅ 2 m ⋅ 1 ( 2 m − 1 ) ) ∑ m = 1 n ( 8 m 3 − 12 m 2 + 6 m − 1 ) = 8 ( n ( n + 1 ) 6 ) 2 − 12 ( n ( n + 1 ) ( 2 n + 1 ) 6 ) + 6 ( n ( n + 1 ) 2 ) − n 2 n 2 ( n + 1 ) 2 − 2 n ( n + 1 ) ( 2 n + 1 ) + 3 n ( n + 1 ) − n n ( n + 1 ) [ 2 n ( n + 1 ) − 2 ( 2 n + 1 ) + 3 ] − n n ( n + 1 ) ( 2 n 2 − 2 n + 1 ) − n n [ ( n + 1 ) ( 2 n 2 − 2 n + 1 ) − 1 ] n ( 2 n 3 − n ) = n 2 ( 2 n 2 − 1 ) \displaystyle \sum^{n}_{m=1}(2m-1)^3 \\ \displaystyle \sum^{n}_{m=1}(8m^3-1-3\cdot 2m \cdot 1(2m-1)) \\ \displaystyle \sum^{n}_{m=1}(8m^3-12m^2+6m-1)=8\left(\frac{n(n+1)}{6}\right)^2-12\left( \frac{n(n+1)(2n+1)}{6}\right)+6\left(\frac{n(n+1)}{2}\right)-n \\ 2n^2(n+1)^2-2n(n+1)(2n+1)+3n(n+1)-n \\ n(n+1)[2n(n+1)-2(2n+1)+3]-n \\ n(n+1)(2n^2-2n+1)-n \\ n[(n+1)(2n^2-2n+1)-1] \\ n(2n^3-n)=\boxed{n^2(2n^2-1)}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

After plodding through all of the calculations, we arrived at the answer.

Is there an alternative interpretation which allows us to find the answer immediately? (No idea if this is true). For example, if we wanted to calculate 1 + 3 + 5 + … + ( 2 n − 1 ) 1 + 3 + 5 + \ldots + (2n-1) , we could compare it to successively building up squares.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...