Answer to Question #122823 in Combinatorics | Number Theory for Ojugbele Daniel

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 "1^2+2^2+3^2+4^2+...+(2n)^2=\\frac{n(2n+1)(4n+1)}{3}"


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

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


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

"1^2+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, "1^2+2^2+3^2+4^2+...+(2n)^2=\\frac{n(2n+1)(4n+1)}{3}."




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