Amazing what you can find in a few seconds using Google. Key words: proof induction sum squares.
Read this (one of 747,000 found in 0.28 seconds): http://math.arizona.edu/~kerl/doc/induction.pdf
Amazing what you can find in a few seconds using Google. Key words: proof induction sum squares.
Read this (one of 747,000 found in 0.28 seconds): http://math.arizona.edu/~kerl/doc/induction.pdf
Mathematical Induction Example 2 --- Sum of Squares
Problem: For any natural number n , 1^2 + 2^2 + ... + n^2 = n( n + 1 )( 2n + 1 )/6.
Proof:
Basis Step: If n = 0, then LHS = 02 = 0, and RHS = 0 * (0 + 1)(2*0 + 1)/6 = 0 .
Hence LHS = RHS.
Induction: Assume that for an arbitrary natural number n,
1^2 + 2^2 + ... + n^2 = n( n + 1 )( 2n + 1 )/6. -------- Induction Hypothesis
To prove this for n+1, first try to express LHS for n+1 in terms of LHS for n, and use the induction hypothesis.
Here let us try
LHS for n + 1 = 1^2 + 2^2 + ... + n^2 + (n + 1)^2 = ( 1^2 + 2^2 + ... + n^2 ) + (n + 1)^2
Using the induction hypothesis, the last expression can be rewritten as
n( n + 1 )( 2n + 1 )/6 + (n + 1)^2
Factoring (n + 1)/6 out, we get
( n + 1 )( n( 2n + 1 ) + 6 ( n + 1 ) )/6
= ( n + 1 )( 2n^2 + 7n + 6 )/6
= ( n + 1 )( n + 2 )( 2n + 3 )/6 ,
which is equal to the RHS for n+1.
Thus LHS = RHS for n+1.
End of Proof.