91Ó°ÊÓ

First, let's check if the given identity holds true for n = 1: \[1^2 = \frac{1}{6}(1)(1+1)(2(1)+1)\] LHS: \(1^2 = 1\) RHS: \(\frac{1}{6}(1)(2)(3) = 1\) Since the LHS and RHS are equal when n = 1, the identity holds true for the base case.

Now, let's assume the given identity is true for n = k, where k is an arbitrary natural number: \[1^2 + 2^2 + \cdots + k^2 = \frac{1}{6}k(k+1)(2k+1)\]

Next, we need to prove that the identity is true for n = k + 1, i.e.: \[1^2 + 2^2 + \cdots + k^2 + (k+1)^2 = \frac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)\] Let's start with the left-hand side: LHS: \(1^2 + 2^2 + \cdots + k^2 + (k+1)^2\) Now, we can replace the sum of squares till k using our assumption: LHS: \[\frac{1}{6}k(k+1)(2k+1) + (k+1)^2\] Now let's factor out (k+1) from the LHS: LHS: \((k+1)\left[\frac{1}{6}k(2k+1) + (k+1)\right]\] Now, we will simplify the expression inside the brackets: \[\frac{1}{6}k(2k+1) + (k+1) = \frac{1}{6}(2k^2 + k) + (k+1)\] \[= \frac{1}{6}(2k^2 + k + 6k + 6)\] \[= \frac{1}{6}(2k^2 + 7k + 6)\] Now, let's work with the right-hand side: RHS: \[\frac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)\] \[= \frac{1}{6}(k+1)(k+2)(2k+3)\] \[= (k+1)\left[\frac{1}{6}(2k^2 + 7k + 6)\right]\] Notice that the LHS and RHS are now equal.

Since the given identity holds true for n = 1 (base case) and it has been proven true for n = k + 1 using the assumption that it is true for n = k, by the principle of mathematical induction, we conclude that the identity: \[1^2 + 2^2 + \cdots + n^2 = \frac{1}{6}n(n+1)(2n+1)\] holds true for all natural numbers n.