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.
On
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)) = $$(x+1)^2 + (x+3)^2\cdot(2x) - (x+1)^2 + (2x+1)^2 + (2x+2)^2\\ \frac 16 (x)(2x+1)(7x+1) + (2x+1)^2 + (2x+2)^2 - (x+1)^2$
by the inductive hypothesis $\frac 16 (x)(2x+1)(7x+1) + (2x+1)^2 + 3(x+1)^2$
$\frac 16 (2x+1) [(x)(7x+1) + 12x+6] + 3(x+1)^2\\ \frac 16 (2x+1) [7x^2+13x +6] + 3(x+1)^2\\ \frac 16 (2x+1) (7x + 6)(x+1) + 3(x+1)^2\\ \frac 16 (x+1)[(2x+1)(7x + 6)+ 18(x+1)]\\ \frac 16 (x+1) [14x^2+37x + 24]\\ \frac 16 (x+1)(2x + 3)(7x + 8)$
$$\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}$$ __