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How can I show that:$$\sum_{k=1}^{n}({k^2})$$ Is equal to: $$\frac{n(n+1)(2n+1)}{6}$$ I know that I would apply the sum formula, should I also be using this formula? $$\sum_{k=1}^{n}k=\frac{n(n+1)}2$$

gt6989b
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Ørangع
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  • "I know that I would apply the sum formula" What is "the sum formula". "should I also be using this formula?" Isn't that formula the sum formula? – fleablood Mar 11 '19 at 19:47

1 Answers1

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Use proof by induction

1 - Demonstrate that it is true for $n=1$

2 - Demonstrate that if it is true for $n=k$ it must also be true for $n=k+1$

This comes down to demonstrating ... $$ (k+1)^2+ \frac{k(k+1)(2k+1)}{6} \\= \frac{(k+1)(k+2)(2(k+1)+1)}{6}$$

WW1
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