20 lines
538 B
TeX
20 lines
538 B
TeX
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\beginsection 1.1
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Prove $1^2+2^2+\cdots+n^2=n(n+1)(2n+1)/6$ for all natural numbers $n$.
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\medskip
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Use mathematical induction.
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First show that proposition $P_1$ is true:
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$$1^2=1(1+1)(2\cdot1+1)/6$$
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Next show that $P_{n+1}$ is true whenever $P_n$ is true:
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$$\eqalign{
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P_n+(n+1)^2&=P_{n+1}\cr
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n(n+1)(2n+1)+(n+1)^2&=(n+1)(n+2)(2n+3)/6\cr
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}$$
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Divide both sides by $(n+1)$ to get
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$$n(2n+1)/6+(n+1)=(n+2)(2n+3)/6$$
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Expand and simplify.
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$$\eqalign{
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(2n^2+n)/6+n+1&=(2n^2+3n+4n+6)/6\cr
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{1\over3}n^2+{7\over6}n+1&={1\over3}n^2+{7\over6}n+1\cr
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}$$
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