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# OK! Now Prove It

First the method using the standard formulae: $$\sum_{i=1}^n i = {1\over 2}n(n + 1)\quad {\rm and} \ \sum_{i=1}^n i^2 = {1\over 6}n(n + 1)(2n + 1).$$

We conjecture that $$\sum _{i=1}^n (2i - 1)^2 = {1 \over 6}(2n -1)2n(2n+1).$$ Consider \begin{eqnarray} \\ \sum_{i=1}^n(2i - 1)^2 &=& 4 \sum_{i=1}^n i^2 - 4\sum_{i=1}^n i + \sum_{i=1}^n 1 \\ &=& 4\left(\frac{1}{6}n(n + 1)(2n + 1)\right ) - 4\left({1\over 2}n(n + 1)\right) + n \\ &=& \frac{1}{6}(2n)(4n^2 + 6n + 2 - 6n - 6 + 3) \\ &=& \frac{1}{6}(2n)(4n^2 - 1) \\ &=& \frac{1}{6}(2n)(2n - 1)(2n + 1) \end{eqnarray}

The second method uses mathematical induction.

The formulae given in the question are easily verified showing that the conjecture is true for $n = 1, 2$ and $3$.

Suppose that the conjecture is true for $n = k$. Then $$1^2 + 3^2 + ... + (2k - 1)^2 = {(2k - 1)(2k)(2k + 1)\over 6}\quad (1)$$ Adding one more term we get \begin{eqnarray} \\ 1^2 + 3^2 + ... + (2k - 1 )^2 + (2k + 1 )^2 &=& \frac{(2k - 1)(2k)(2k + 1)}{6} + (2k + 1)^2 \\ &=& \frac{(2k - 1)(2k)(2k + 1) + 6(2k + 1)^2}{6} \\ &=& \frac{(2k + 1)(4k^2 + 10k + 6)}{6} \\ &=& \frac{(2k + 1)(2k + 2)(2k + 3)}{6}. \end{eqnarray} This is essentially the same as (1) but here $k$ is replaced by $k + 1$. Thus if the conjecture is true for $n = k$, it is also true for $n = k + 1$.

Thus by the axiom of induction $$1^2 + 3^2 + ... + (2n - 1)^2 = {(2n - 1)(2n)(2n + 1)\over 6}.$$

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Congratulations to Phong Quach of Debney Park Secondary College, Tomas from Malmesbury School, Guildford, Chor Kiang Tan, Vassil from Lawnswood High School, Tom from Madras College, Adam from King James's School, Knaresborough, Alex from King Edward and Queen Mary School, Lytham and Andaleeb from Woodhouse Sixth Form College, London who all proved this result using mathematical induction. Congratulations also to Daniel, and Alex, also to Mark and Eduardo from the British School of Manila and to Michael and Sue of Madras College and Yiwen of the Chines High School, Singapore for their alternative method using the standard formulae .

First the method using the standard formulae: $$\sum_{i=1}^n i = {1\over 2}n(n + 1)\quad {\rm and} \ \sum_{i=1}^n i^2 = {1\over 6}n(n + 1)(2n + 1).$$

We conjecture that $$\sum _{i=1}^n (2i - 1)^2 = {1 \over 6}(2n -1)2n(2n+1).$$ Consider \begin{eqnarray} \\ \sum_{i=1}^n(2i - 1)^2 &=& 4 \sum_{i=1}^n i^2 - 4\sum_{i=1}^n i + \sum_{i=1}^n 1 \\ &=& 4\left(\frac{1}{6}n(n + 1)(2n + 1)\right ) - 4\left({1\over 2}n(n + 1)\right) + n \\ &=& \frac{1}{6}(2n)(4n^2 + 6n + 2 - 6n - 6 + 3) \\ &=& \frac{1}{6}(2n)(4n^2 - 1) \\ &=& \frac{1}{6}(2n)(2n - 1)(2n + 1) \end{eqnarray}

The second method uses mathematical induction.

The formulae given in the question are easily verified showing that the conjecture is true for $n = 1, 2$ and $3$.

Suppose that the conjecture is true for $n = k$. Then $$1^2 + 3^2 + ... + (2k - 1)^2 = {(2k - 1)(2k)(2k + 1)\over 6}\quad (1)$$ Adding one more term we get \begin{eqnarray} \\ 1^2 + 3^2 + ... + (2k - 1 )^2 + (2k + 1 )^2 &=& \frac{(2k - 1)(2k)(2k + 1)}{6} + (2k + 1)^2 \\ &=& \frac{(2k - 1)(2k)(2k + 1) + 6(2k + 1)^2}{6} \\ &=& \frac{(2k + 1)(4k^2 + 10k + 6)}{6} \\ &=& \frac{(2k + 1)(2k + 2)(2k + 3)}{6}. \end{eqnarray} This is essentially the same as (1) but here $k$ is replaced by $k + 1$. Thus if the conjecture is true for $n = k$, it is also true for $n = k + 1$.

Thus by the axiom of induction $$1^2 + 3^2 + ... + (2n - 1)^2 = {(2n - 1)(2n)(2n + 1)\over 6}.$$

Find $S_r = 1^r + 2^r + 3^r + ... + n^r$ where r is any fixed positive integer in terms of $S_1, S_2, ... S_{r-1}$.

Bricks are 20cm long and 10cm high. How high could an arch be built without mortar on a flat horizontal surface, to overhang by 1 metre? How big an overhang is it possible to make like this?