Proof of Series SumDate: 06/19/97 at 01:16:59 From: piggy Subject: mathematics How do you prove: 1x2 + 2x3 + 3x4 + 4x5 + 5x6 + 6x7......+n(n+1) = (nx(n+1) x (n+2))/3? Date: 06/19/97 at 05:05:52 From: Doctor Anthony Subject: Re: mathematics Use SIGMA(1 to n)[r(r+1)] SIGMA[r^2 + r] = n(n+1)(2n+1)/6 + n(n+1)/2 = [n(n+1)/6][2n+1 + 3] = [n(n+1)/6][2n+4] = [n(n+1)(n+2)]/3 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 06/28/97 at 12:09:11 From: peggy Subject: Re: mathematics Dr. Math, Thank you for your response. However, I still do not understand how to prove it. Can you explain it in details? (The solution uses SIGMA. I don't know what this is.) Date: 06/28/97 at 18:16:15 From: Doctor Anthony Subject: Re: mathematics I was using the standard formulae for the sum 1^2 + 2^2 + 3^2 + .. + n^2 and for 1 + 2 + 3 + ... + n. However if you are not familiar with these we can prove the result as follows. Consider n(n+1)(n+2) - (n-1)n(n+1) = n(n+1)[n+2-n+1] = 3n(n+1) So we now consider a series of equations as follows: n(n+1)(n+2) - (n-1)n(n+1) = 3n(n+1) (n-1)n(n+1) - (n-2)(n-1)(n) = 3(n-1)n (n-2)(n-1)n - (n-3)(n-2)(n-1) = 3(n-2)(n-1) ----------------------------- ----------- ----------------------------- ----------- 3 x 4 x 5 - 2 x 3 x 4 = 3(3)(4) 2 x 3 x 4 - 1 x 2 x 3 = 3(2)(3) Add all the 1 x 2 x 3 - 0 x 1 x 2 = 3(1)(2) equations ----------------------------------------------- n(n+1)(n+2) - 0 = 3[1x2 + 2x3 + ... + n(n+1)] Note that when adding all the equations there is a cancelling of terms on the lefthand side between adjacent lines. The term -(n-1)n(n+1) in the first line cancels with the term +(n-1)n(n+1) in the second line, and so on down all the equations. The only terms that do not cancel are the first term in the top equation and the second term in the last equation. n(n+1)(n+2) We then get ----------- = 1x2 + 2x3 + 3x4 + .... + n(n+1) 3 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/