Question 1104787
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The formula to be proved is<br>
1*2+3*4+5*6...+(2n-1)*(2n)=(1/3)n(n+1)(4n-1)<br>
Step 1: show the formula is true for n=1.
{{{1*2 = (1/3)(1)(2)(3)}}}
{{{2 = 2}}}
Yes, the formula is true for n=1.<br>
Step 2: Assume the formula is true for n=k and show that it follows that it is also true for n=k+1.<br>
{{{1*2+3*4+5*6}}} ... +{{{(2k-1)*(2k) + (2k+1)(2k+2)}}} =
{{{((1/3)k(k+1)(4k-1)) + (2k+1)(2k+2)}}} =
{{{((1/3)k(k+1)(4k-1)) + 2(2k+1)(k+1)}}} =
{{{(k+1)((1/3)k(4k-1)+4k+2)}}} =
{{{(k+1)((1/3)k(4k-1)+(1/3)(12k+6))}}} =
{{{(k+1)((1/3)(4k^2-k+12k+6))}}} =
{{{(1/3)(k+1)(4k^2+11k+6)}}} =
{{{(1/3)(k+1)(k+2)(4k+3)}}} =
{{{(1/3)(k+1)((k+1)+1)(4((k+1)-1))}}}<br>
This last expression is the formula we were to prove, with "k" replaced by "k+1", so the proof my mathematical induction is complete.<br>
Note when doing a proof like this, it can be hard to see where to go with the algebraic manipulations.<br>
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.<br>
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.