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To prove:

$$(x+1)^2+(x+2)^2...(2x)^2=(x(2x+1)(7x+1))/6$$

I got to the proving part using $P(K+1)$ which comes out to be $(x(2x+1)(7x+1))/6 - (x+1)^2 + (2x+2)^2$ not sure if my logic is right, what should i do after this? i tried to distribute and set a common denominator but still cant figure it out.

abiessu
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James Chen
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  • What is $n$? Distribution and recombination is correct... – abiessu Nov 07 '16 at 23:35
  • sorry, should have been x not n – James Chen Nov 07 '16 at 23:36
  • Fixed. So when you set the denominator at $6$ and fully distribute and recombine terms, what do you get? – abiessu Nov 07 '16 at 23:39
  • Your $P(k+1)$ is not correct. – hamam_Abdallah Nov 07 '16 at 23:41
  • (14n^3+15n^2++13n+18)/6 – James Chen Nov 07 '16 at 23:45
  • How is my P(K+1) incorrect? my logic was p(k+1)will be (x+2)^2+(x+2)^2+....+(2n+2)^2 you would need to take out the (x+1)^2 from the equation – James Chen Nov 07 '16 at 23:47
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    Your use of some of the notation is slightly strange, but still understandable. (It is unusual, for example, to use $x$ as a variable whose value is only ever an integer.) Anyway, you do subtract $(x+1)^2$, as you claimed, but must add in both $(2x+1)^2$ and $(2x+2)^2$, not just $(2x+2)^2$. – Dylan Nov 07 '16 at 23:55
  • hum, where did you get (2x+1)^2 from? – James Chen Nov 08 '16 at 00:05
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    We're adding all of the consecutive terms, not just the even ones, right? So after $2x$ comes $2x+1$. You don't jump straight to $2x+2$. Try putting in actual numbers in the place of $x$ and seeing what you get. – Dylan Nov 08 '16 at 00:20

2 Answers2

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$$\begin{align} \sum_{r={x+1}}^{2x} r^2 &=\sum_{r=1}^{2x}r^2-\sum_{r=1}^x r^2\\ &=\frac 16 (2x)(2x+1)(4x+1)-\frac 16x(x+1)(2x+1)\\ &=\frac 16 x(2x+1)\left[2(4x+1)-(x+1)\right]\\ &=\frac 16 x(2x+1)(7x+1)\end{align}$$ __

Hypergeometricx
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base case x = 1 $(2)^2 = \frac 16 (1)(3)(8)\\ 4 = 4$

Inducuctive hypothesis:

Suppose $(x+1)^2 + (x+2)^2\cdot(2x) = \frac 16 (x)(2x+1)(7x+1)$

Using the inductive hypothesis, we will show that:

$(x+2)^2 + (x+3)^2\cdot(2(x+1)) = \frac 16 (x+1)(2(x+1)+1)(7(x+1)+1)$

$(x+2)^2 + (x+3)^2\cdot(2(x+1)) = $$(x+1)^2 + (x+3)^2\cdot(2x) - (x+1)^2 + (2x+1)^2 + (2x+2)^2\\ \frac 16 (x)(2x+1)(7x+1) + (2x+1)^2 + (2x+2)^2 - (x+1)^2$

by the inductive hypothesis $\frac 16 (x)(2x+1)(7x+1) + (2x+1)^2 + 3(x+1)^2$

$\frac 16 (2x+1) [(x)(7x+1) + 12x+6] + 3(x+1)^2\\ \frac 16 (2x+1) [7x^2+13x +6] + 3(x+1)^2\\ \frac 16 (2x+1) (7x + 6)(x+1) + 3(x+1)^2\\ \frac 16 (x+1)[(2x+1)(7x + 6)+ 18(x+1)]\\ \frac 16 (x+1) [14x^2+37x + 24]\\ \frac 16 (x+1)(2x + 3)(7x + 8)$

Doug M
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