The given expression is a sum of squares of consecutive natural numbers. We can prove the formula for the sum of squares by mathematical induction.
Base Case (n = 1):
\[1^2 = \frac{1(1+1)(2 \cdot 1 + 1)}{6} = \frac{6}{6} = 1.\]
The formula holds for n = 1.
Inductive Step:
Assume that the formula holds for some arbitrary positive integer k, i.e.,
\[1^2 + 2^2 + 3^2 + \ldots + k^2 = \frac{k(k+1)(2k+1)}{6}.\]
Now, let's prove it for (k+1):
\[1^2 + 2^2 + 3^2 + \ldots + k^2 + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2.\]
We can rewrite the denominator to have a common denominator:
\[= \frac{k(k+1)(2k+1) + 6(k+1)^2}{6}.\]
Factor out (k+1) from the numerator:
\[= \frac{(k+1)[k(2k+1) + 6(k+1)]}{6}.\]
Simplify the expression in the brackets:
\[= \frac{(k+1)[2k^2 + 7k + 6]}{6}.\]
Factor the quadratic expression in the brackets:
\[= \frac{(k+1)(k+2)(2k+3)}{6}.\]
This matches the form of the formula for n = k+1:
\[\frac{(k+1)[(k+1)+1][2(k+1)+1]}{6}.\]
So, by mathematical induction, the formula holds for all positive integers n.
0 टिप्पणियाँ:
Post a Comment