Mathematical Induction Problem (Grade 11)

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Prove my M.I. that for all positive integers n,
1x3^2 + 2x5^2 + 3x7^2 + ... + n(2n+1)^2 = [n(n+1)(6n^2+14n+7)]/6

My workings:
When n = 1, L.H.S. = 9, R.H.S. = [1(2)(27)]/6 = 9
L.H.S. = R.H.S. = 9
S(1) is true
Assume that S(k) is true
i.e. 1x3^2 + 2x5^2 + 3x7^2 + ... + k(2k+1)^2 = [k(k+1)(6k^2+14k+7)]/6
For n = k+1
1x3^2 + 2x5^2 + 3x7^2 + ... + k(2k+1)^2 + (k+1)(2k+3)^2
= [k(k+1)(6k^2+14k+7)]/6 + (k+1)(2k+3)^2
= 1/6 (k+1) [k(6k^2+14k+7) + 6(2k+3)^2)

Can anyone teach me to finish this question?
I encounter problems when I tried to finish this type of questions

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Its perfect where you have left it

Simplify and you will get ur answer

LHS = 1/6 (k+1) [k(6k^2+14k+7) + 6(2k+3)^2)
= 1/6 (k+1) [6k^3+14k^2+7k+6(4k^2+12k+9)]
= 1/6 (k+1) [6k^3+14k^2+7k+24k^2+72k+54]
= 1/6 (k+1) [6k^3+38k^2+79k+54]

Now see if you can get (k+2) out of [6k^3+38k^2+79k+54] using polynomial factorization method ....

Code:

   _____________________

2 | 6   38    79    54

  |    -12   -54   -54

--------------------------

  | 6   26    27    00

YES u could ....

so (k+2) (6k^2+26k+27) = [6k^3+38k^2+79k+54]

Now see if you can get represent (6k^2+26k+27) in terms of (k+1)

6k^2+26k+27 = 6(k+1)^2+14(k+1)+7

So finally

LHS = ((k+1)(k+2)[6(k+1)^2+14(k+1)+7])/6

You are done!

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Mathematical Induction Problem (Grade 11)

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