SOLUTION: use the principals of mathematical induction to prove the following statement 1*2+3*4+5*6...+(2n-1)*(2n)=(1/3)n(n+1)(4n-1)

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Question 1104787: use the principals of mathematical induction to prove the following statement 1*2+3*4+5*6...+(2n-1)*(2n)=(1/3)n(n+1)(4n-1)
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The formula to be proved is

1*2+3*4+5*6...+(2n-1)*(2n)=(1/3)n(n+1)(4n-1)

Step 1: show the formula is true for n=1.
1%2A2+=+%281%2F3%29%281%29%282%29%283%29
2+=+2
Yes, the formula is true for n=1.

Step 2: Assume the formula is true for n=k and show that it follows that it is also true for n=k+1.

1%2A2%2B3%2A4%2B5%2A6 ... +%282k-1%29%2A%282k%29+%2B+%282k%2B1%29%282k%2B2%29 =
%28%281%2F3%29k%28k%2B1%29%284k-1%29%29+%2B+%282k%2B1%29%282k%2B2%29 =
%28%281%2F3%29k%28k%2B1%29%284k-1%29%29+%2B+2%282k%2B1%29%28k%2B1%29 =
%28k%2B1%29%28%281%2F3%29k%284k-1%29%2B4k%2B2%29 =
%28k%2B1%29%28%281%2F3%29k%284k-1%29%2B%281%2F3%29%2812k%2B6%29%29 =
%28k%2B1%29%28%281%2F3%29%284k%5E2-k%2B12k%2B6%29%29 =
%281%2F3%29%28k%2B1%29%284k%5E2%2B11k%2B6%29 =
%281%2F3%29%28k%2B1%29%28k%2B2%29%284k%2B3%29 =
%281%2F3%29%28k%2B1%29%28%28k%2B1%29%2B1%29%284%28%28k%2B1%29-1%29%29

This last expression is the formula we were to prove, with "k" replaced by "k+1", so the proof my mathematical induction is complete.

Note when doing a proof like this, it can be hard to see where to go with the algebraic manipulations.

The key to seeing which direction to go can be seen in the third line of work above, where I factored "(2k+2)" into "2(k+1)". I did this because I knew that, with a factor "k" in the formula for n=k, I would need a factor of "(k+1)" in the formula for n = k+1.

Once I then factored out that common factor of (k+1), it was easy to see what I had to do with the rest of the expression.