(2n%2B1)%2F6-proof.jpeg "12+22+32+...+n2 = N(n+1)(2n+1)/6 Proof")
Proof of 12+22+32+...+n2 = n(n+1)(2n+1)/6
In mathematics, the formula 12+22+32+...+n2 = n(n+1)(2n+1)/6 is a well-known identity that has been widely used in various mathematical contexts. In this article, we will provide a proof of this formula using mathematical induction.
The Formula
The formula states that the sum of the squares of the first n natural numbers is given by:
1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6
Base Case
To prove this formula, we will use mathematical induction. We will start by verifying the base case, which is the case where n = 1.
1^2 = 1(1+1)(2*1+1)/6 = 1(2)(3)/6 = 1
This shows that the formula is true for n = 1.
Inductive Step
Now, we will assume that the formula is true for some natural number k, and we will prove that it is also true for k+1.
Inductive Hypothesis
Assume that:
1^2 + 2^2 + 3^2 + ... + k^2 = k(k+1)(2k+1)/6
Inductive Conclusion
We need to show that:
1^2 + 2^2 + 3^2 + ... + k^2 + (k+1)^2 = (k+1)((k+1)+1)(2(k+1)+1)/6
Using the inductive hypothesis, we can rewrite the left-hand side as:
k(k+1)(2k+1)/6 + (k+1)^2
= k(k+1)(2k+1)/6 + (k+1)(k+1)
= (k+1)(k(2k+1)/6 + k+1)
= (k+1)((2k^2 + k + 6)/6 + k+1)
= (k+1)((2k^2 + 7k + 6)/6)
= (k+1)((k+2)(2k+3)/6)
= (k+1)((k+1)+1)(2(k+1)+1)/6
This shows that the formula is true for k+1, and therefore it is true for all natural numbers n.
Conclusion
In conclusion, we have provided a proof of the formula 12+22+32+...+n2 = n(n+1)(2n+1)/6 using mathematical induction. This formula is a fundamental identity in mathematics, and it has numerous applications in various mathematical contexts.
