Oct 23, 2013 Thread starter #1 K KOO New member Oct 19, 2013 19 Prove that for all nEN 1^2 + 3^2 + 5^2 + .... + (2n-1)^2 = (4n^3 - n) / 3 My Solution) If n = 1, 4(1)^3 - 1 / 3 = 1 so base case holds. Assume 1^2 + 3^2 + .... + (2k-1)^2 = (4k^3 - k) / 3 What next?
Prove that for all nEN 1^2 + 3^2 + 5^2 + .... + (2n-1)^2 = (4n^3 - n) / 3 My Solution) If n = 1, 4(1)^3 - 1 / 3 = 1 so base case holds. Assume 1^2 + 3^2 + .... + (2k-1)^2 = (4k^3 - k) / 3 What next?
Oct 23, 2013 #2 Deveno Well-known member MHB Math Scholar Feb 15, 2012 1,967 Well, then, obviously: $1^2 + 3^2 + 5^2 + \cdots + (2k - 1)^2 + (2k + 1)^2 = \dfrac{4k^3 - k}{3} + (2k + 1)^2$ $= \dfrac{4k^3 - k}{3} + \dfrac{12k^2 + 12k + 3}{3}$ $= \dfrac{4k^3 + 12k^2 + 12k + 4 - k - 1}{3}$ $= \dfrac{4(k+1)^3 - (k+1)}{3}$
Well, then, obviously: $1^2 + 3^2 + 5^2 + \cdots + (2k - 1)^2 + (2k + 1)^2 = \dfrac{4k^3 - k}{3} + (2k + 1)^2$ $= \dfrac{4k^3 - k}{3} + \dfrac{12k^2 + 12k + 3}{3}$ $= \dfrac{4k^3 + 12k^2 + 12k + 4 - k - 1}{3}$ $= \dfrac{4(k+1)^3 - (k+1)}{3}$