(1+2+3+.......+(n-1))^2 + n^3

= (n(n-1)/2)^2 + n^3

= (n^4 -2n^3 +n^2)/4 + n^3

= (n^4 +2n^3 +n^2)/4

=n^2*(n+1)^2/4

= (n(n+1)/2)^2

= (1+2+3+.......+n)^2

yes it is true

Solved

Posted on 2011-02-11

Hello

I have seen somewhere that this is true, can somebody help me in proving this?

Sham

I have seen somewhere that this is true, can somebody help me in proving this?

Sham

3 Comments

(1+2+3+.......+(n-1))^2 + n^3

= (n(n-1)/2)^2 + n^3

= (n^4 -2n^3 +n^2)/4 + n^3

= (n^4 +2n^3 +n^2)/4

=n^2*(n+1)^2/4

= (n(n+1)/2)^2

= (1+2+3+.......+n)^2

yes it is true

And k = (n^2 - n)/2.

Left hand side (lhs) = k^2 +n^3.

lhs = ((n^2 - n)/2)^2 + n^3.

lhs = (n^4 - 2n^3 + n^2)/4 + n^3.

lhs = (n^4 - 2n^3 + n^2 + 4n^3)/4.

lhs = (n^4 + 2n^3 + n^2)/4.

Right hand side (rhs) = (k + n)^2

rhs = ((n^2 - n)/2 + n)^2.

rhs = ((n^2 - n + 2n)/2)^2.

rhs = ((n^2 + n)/2)^2.

rhs = (n^4 + 2n^3 + n^2)/4.

Thus, lhs = rhs.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Title | # Comments | Views | Activity |
---|---|---|---|

How to solve this equation | 3 | 51 | |

Probability | 2 | 52 | |

Graph function | 4 | 72 | |

Windows Batch File - Count Down | 4 | 61 |

Join the community of 500,000 technology professionals and ask your questions.

Connect with top rated Experts

**19** Experts available now in Live!