INDUCTION HELP (1 Viewer)

SMARTYPANTS_SINGH · Oct 12, 2023

Idk if I'm being special but I literally get so close but don't get the proof in the end. Can someone help me please I've spent way more time on this than I feel like I should have. Don't worry about the base case and the assumption I've got that.

SMARTYPANTS_SINGH · Oct 12, 2023

nvm just got it

Luukas.2 · Oct 13, 2023

This result can be proved by summing APs without induction, considering the cases where $n$ is odd and even separately:

$\begin{align*} 1^2 - 2^2 + 3^2 - 4^2 + ... + (-1)^{n - 1}n^2 &= 1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + ... + (n-1)^2 - n^2 \qquad \text{supposing $n$ is even} \\ &= (1 + 2)(1 - 2) + (3 + 4)(3 - 4) + (5 + 6)(5 - 6) + ... + [(n - 1) + n][(n - 1) - n] \\ &= 3 \times -1 + 7 \times -1 + 11 \times -1 + ... + (2n - 1) \times -1 \\ &= -\big[3 + 7 + 11 + ... + (2n - 1)\big] \\ &= -\frac{1}{2} \times \frac{n}{2}\big[3 + (2n - 1)\big] \qquad \text{on summing the AP, which has $\frac{n}{2}$ terms} \\ &= -\frac{n \times 2(n + 1)}{4} \\ &= (-1)^{n - 1}\frac{n(n + 1)}{2} \qquad \text{as $(-1)^{n - 1} = -1$ when $n$ is even} \end{align*}$

$\begin{align*} 1^2 - 2^2 + 3^2 - 4^2 + ... + (-1)^{n - 1}n^2 &= 1^2 - 2^2 + 3^2 - 4^2 + ... + (n - 2)^2 - (n-1)^2 + n^2 \qquad \text{supposing $n$ is odd} \\ &= (1 + 2)(1 - 2) + (3 + 4)(3 - 4) + ... + [(n - 2) + (n - 1)][(n - 2) - (n - 1)] + n^2 \\ &= -\big[3 + 7 + 11 + ... + (2n - 3)\big] + n^2 \\ &= n^2 - \frac{1}{2} \times \frac{n - 1}{2}\big[3 + (2n - 3)\big] \qquad \text{on summing the AP, which has $\frac{n - 1}{2}$ terms} \\ &= n^2 - \frac{(n - 1) \times 2n}{4} \\ &= \frac{2n^2 - n(n - 1)}{2} \\ &= \frac{n(n + 1)}{2} \\&= (-1)^{n - 1}\frac{n(n + 1)}{2} \qquad \text{as $(-1)^{n - 1} = 1$ when $n$ is odd} \end{align*}$

INDUCTION HELP (1 Viewer)

SMARTYPANTS_SINGH

NESA WHY?

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SMARTYPANTS_SINGH

NESA WHY?

Luukas.2

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