Prove that the formula 1⋅3+2⋅4+3⋅5+...+n(n+2)=n(n+1)(2n+7)6 is true for all natural
numbers n.
By using mathematical induction,
Let P(n) denote the statement 1⋅3+2⋅4+3⋅5+...+n(n+2)=n(n+1)(2n+7)6.
Then, we need to show that P(1) is true. So,
1⋅3=(1)(1+1)(2(1)+7)63=(2)(9)63=1863=3
Thus, we prove the first principle of the mathematical induction. More
over, assuming that P(k) is true, then
1⋅3+2⋅4+3⋅5+...k(k+2)=k(k+1)(2k+7)6
Now, by showing P(k+1), we have
1⋅3+2⋅4+3⋅5+...k(k+2)+(k+1)[(k+1)+2]=(k+1)[(k+1)+1][2(k+1)+7]61⋅3+2⋅4+3⋅5+...k(k+2)+(k+1)(k+3)=(k+1)(k+2)(2k+9)61⋅3+2⋅4+3⋅5+...k(k+2)+(k+1)(k+3)=(k2+3k+2)(2k+9)61⋅3+2⋅4+3⋅5+...k(k+2)+(k+1)(k+3)=2k3+9k2+6k2+27k+4k+1861⋅3+2⋅4+3⋅5+...k(k+2)+(k+1)(k+3)=2k3+15k2+31k+1861⋅3+2⋅4+3⋅5+...k(k+2)+(k+1)(k+3)=13k3+52k2+316k+3
We start with the left side and use the induction hypothesis to obtain
the right side of the equation:
=[1⋅3+2⋅4+3⋅5+k(k+2)]+[(k+1)(k+3)]Group the first k terms=k(k+1)(2k+7)6+(k+1)(k+3)Induction hypothesis=2k3+9k2+7k6Expand=13k3+32k2+76k+k2+4k+3Simplify=13k3+52k2+316k+3
Therefore, P(k+1) follows from P(k), and this completes the
induction step.
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