- #1
iainfs
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Homework Statement
Prove
[tex]\sum^{2n}_{r=1} r^3 = n^2(2n+1)^2[/tex]
by mathematical induction.
My proof is below but it is rather long winded. Is there a quicker way of doing the algebraic manipulation? Preferably one that forgoes the polynomial division and multiplying out all the brackets!
The Attempt at a Solution
Base case, [tex]n=1[/tex]:
[tex]\sum^{2 \times 1}_{r=1} r^3 = 1^2 (2 \times 1+1)^2 = 1 \times 9 = 9[/tex]
[tex]= 1^3 + 2^3 = 1 + 8 = 9[/tex]
Inductive step, assume true for [tex]n=k, \quad k \in \mathbb{N}[/tex]
[tex]\sum^{2k}_{r=1} r^3 = k^2(2k+1)^2[/tex]
Adding the next two terms, [tex]n=k+1[/tex]
[tex]\sum^{2(k+1)}_{r=1} r^3 = k^2(2k+1)^2 + (2k+1)^3 + (2k+2)^3[/tex]
Multiplying out and collecting terms
[tex]\sum^{2k+2}_{r=1} r^3 = 4k^4 + 20k^3 + 37k^2 + 30k + 9[/tex]
By factor theorem, (k+1) is a factor.
By long-hand polynomial division
[tex]= (k+1)(4k^3 + 16k^2 + 21k + 9)[/tex]
By factor theorem, (2k+3) is a factor.
By long-hand polynomial division
[tex]= (k+1)(2k+3)(2k^2 + 5k + 3)[/tex]
Factorising the quadratic
[tex]= (k+1)(2k+3)(2k+3)(k+1)[/tex]
[tex]= (k+1)^2(2k+3)^2[/tex]
[tex]= (k+1)^2(2(k+1)+1)^2[/tex]
So if the statement is true for [tex]n=k[/tex], it is true for [tex]n=k+1[/tex].
Therefore
[tex]\sum^{2n}_{r=1} r^3 = n^2(2n+1)^2[/tex]
for all [tex] n \geq 1, \quad n \in \mathbb{N}[/tex]
Thanks for any suggestions on making this simpler.