Answer to Question #231396 in Discrete Mathematics for citra

Question #231396

Provethat12 +32 +52 +⋯+(2n+1)2 = (n+1)(2n+1)(2+ 3)∕3 whenever is a nonnegative integer.


1
Expert's answer
2021-08-31T12:55:57-0400

Let us prove that 12+32+52++(2n+1)2=(n+1)(2n+1)(2n+3)31^2 +3^2 +5^2 +⋯+(2n+1)^2 = \frac{(n+1)(2n+1)(2n + 3)}{3} using method of mathematical induction.


For n=0n=0 the left side is equal to 11=1,1^1=1, and the right side is equal to (0+1)(20+1)(20+3)333=1.\frac{(0+1)(2\cdot 0+1)(2\cdot 0 + 3)}{3}\frac{3}{3}=1.


Let for n=kn=k the formula 12+32+52++(2k+1)2=(k+1)(2k+1)(2k+3)31^2 +3^2 +5^2 +⋯+(2k+1)^2 = \frac{(k+1)(2k+1)(2k + 3)}{3} is true.


Let us prove for n=k+1:n=k+1:


12+32+52++(2k+1)2+(2(k+1)+1)2=(k+1)(2k+1)(2k+3)3+(2k+3)21^2 +3^2 +5^2 +⋯+(2k+1)^2+(2(k+1)+1)^2 = \frac{(k+1)(2k+1)(2k + 3)}{3}+(2k+3)^2


=(2k+3)((2k2+3k+13+(2k+3))=(2k+3)2k2+3k+1+6k+93=(2k+3)2k2+9k+103= (2k+3)(\frac{(2k^2+3k+1}{3}+(2k+3))=(2k+3)\frac{2k^2+3k+1+6k+9}{3}=(2k+3)\frac{2k^2+9k+10}{3}


=(2k+3)(k+2)(2k+5)3=(k+2)(2k+3)(2k+5)3=((k+1)+1)(2(k+1)+1)(2(k+1)+3)3.=(2k+3)\frac{(k+2)(2k+5)}{3}=\frac{(k+2)(2k+3)(2k+5)}{3}=\frac{((k+1)+1)(2(k+1)+1)(2(k+1)+3)}{3}.


We conclude that this formula is true for any non-negative integer number n.n.


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