How do you prove the 1^2 +2^2+.....+n^2 = n/6 (n+1) (2n+1) by induction?

There are three simple steps to prove anything by induction.

Step 1: When n=1

12=1

then: 1/6 x 1x (1+1) x (2+1)

=1x2x3/6 =1

Therefore it is true for n=1 

Step 2: Assume it is true for n=k

so, 

1+ 22+....+k= k/6 (k+1) (2k+1) 

Step 3: Solve for n= k+1

12+22+.....+ k2 + (k+1)= k/6 (k+1) (2k+1) + (k+1)2

= (k+1)/6 (k(2k+1) + 6(k+1) )

= (k+1)/6 (2k2+7k+6)

= (k+1)/6 x (2k+3) x (k+2)

= (k+1)/6 ( (k+1)+1 ) (2(k+1)+1) )

Therefore it is true for n= k+1 

Thererore it must be true for all values. 

VI

Related Maths A Level answers

All answers ▸

Find exact solution to 2ln(2x+1) - 10 =0


Given that log3 (c ) = m and log27 (d )= n , express c /(d^1/2) in the form 3^y, where y is an expression in terms of m and n.


Find the max/min value of the function: f(x) = 5x^2 - 20x + 15


Find the equation for the tangent to the curve y^3 + x^3 + 3x^2 + 2y + 8 = 0 at the point (2,1)