Question

Prove by induction that the sum of any six consecutive squares leaves a remainder of seven when divided by 12.

Answer 1

Let us assume six consecutive nos. re x, x+1, x+2, x+3, x+4, x+5

Then, sum of their squares, 

S = $x^{2}$ +$(x+1)^{2}$ +$(x+2)^{2}$ +$(x+3)^{2}$ + $(x+4)^{2}$ + $(x+5)^{2}$

   = $x^{2}$ +( $x^{2}$ +1+ 2x ) + ($x^{2}$ +4 +4x) + ($x^{2}$ +9 +6x) + ($x^{2}$ + 16 +8x) + ($x^{2}$ +25 +10x)

   = 6$x^{2}$ +30x +55

When S is divided by 12 i.e. $\frac{6x^{2} + 30x + 55}{12}$

quotient =  0.5$x^{2}$ + 2.5x +4

remainder = 7

Hence, it is proved that sum of any six consecutive squares leaves a remainder of 7 when divided by 12.