Question #122823
Find the sum of the series 1^2 + 2^2 + 3^2 + 4^2 +.... Up to 2n terms.
1
Expert's answer
2020-06-23T14:50:39-0400

We will use the Principle of Mathematical Induction to show that 12+22+32+42+...+(2n)2=n(2n+1)(4n+1)31^2+2^2+3^2+4^2+...+(2n)^2=\frac{n(2n+1)(4n+1)}{3}


1) Show that it is true for n=1:n=1: 12+22=5=1×3×531^2+2^2=5=\frac{1\times 3 \times 5}{3}

2) Assume it is true for n=k:n=k: 12+22+...+(2k)2=k(2k+1)(4k+1)31^2+2^2+...+(2k)^2=\frac{k(2k+1)(4k+1)}{3}


Now, prove it is true for k+1:k+1:

12+22+...+(2k+2)2=(12+...+(2k)2)+(2k+1)2+(2k+2)2=k(2k+1)(4k+1)3+(2k+1)2+(2k+2)2=k(2k+1)(4k+1)+3(2k+1)2+12(k+1)23=8k3+30k2+37k+153=(k+1)(2(k+1)+1)(4(k+1)+1)31^2+2^2+...+(2k+2)^2=(1^2+...+(2k)^2)+(2k+1)^2+(2k+2)^2= \frac{k(2k+1)(4k+1)}{3}+(2k+1)^2+(2k+2)^2= \frac{k(2k+1)(4k+1)+3(2k+1)^2+12(k+1)^2}{3}=\frac{8k^3+30k^2+37k+15}{3}=\frac{(k+1)(2(k+1)+1)(4(k+1)+1)}{3}

It is true.

So, 12+22+32+42+...+(2n)2=n(2n+1)(4n+1)3.1^2+2^2+3^2+4^2+...+(2n)^2=\frac{n(2n+1)(4n+1)}{3}.




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