Prove that the formula 12+22+32+...+n2=n(n+1)(2n+1)6 is true for all natural numbers n.
By using mathematical induction,
Let P(n) denote the statement 12+22+32+...+n2=n(n+1)(2n+1)6
So, P(1)=4(1+1)(2(1)+1)6=(2)(3)6=66=1. Thus, we prove the first principle of the mathematical induction.
More over, assuming that P(k) is true, then 12+22+32+...k2=k(k+1)(2k+1)6
Now, by showing P(k+1), we have
12+22+32+...k2+(k+1)2=(k+1)[(k+1)+1][2(k+1)+1]612+22+32+...k2+(k+1)2=(k+1)(k+2)(2k+3)612+22+32+...k2+(k+1)2=(k2+3k+2)(2k+3)612+22+32+...k2+(k+1)2=2k3+6k2+4k+3k2+9k+6612+22+32+...k2+(k+1)2=2k3+9k2+13k+6612+22+32+...k2+(k+1)2=13k3+32k2+136k+1
We start with the left side and use the induction hypothesis to obtain the right side of the equation:
=[12+22+32+k2]+[(k+1)2]Group the first k terms=k(k+1)(2k+1)6+(k+1)2Induction hypothesis=2k3+3k2+k6+k2+2k+1Expand=13k3+12k2+16k+k2+2k+1=13k3+32k2+136k+1
Thus, P(k+1) follows from P(k), and this completes the induction step.
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