Question

If k1,k2,k3.........kn are odd natural no.then the remainder when k1 ^2+k2^2+.....kn^2 is divided by 4 is always equal to

Answer 1

Answer:

when ${k_1}^2+{k_2}^2+{k_3}^2+..........+{k_n}^2$ is divided by 4, the remainder is $k_1+k_2+k_3+......+k_n$

Step-by-step explanation:

Given:

$k_1,k_2,k_3..........k_n$ are odd natrual numbers.

Then,

$k_1=2x_1+1$

$k_2=2x_2+1$

$k_3=2x_3+1$

.

.

.

$k_n=2x_n+1$

where $x_1,x_2.......x_n$ are integers

Now,

${k_1}^2+{k_2}^2+{k_3}^2+..........+{k_n}^2$

$=(2x_1+1)^2+(2x_2+1)^2+(2x_3+1)^2+..........+(2x_n+1)^2$

$=(4{x_1}^2+2x_1+1)+(4{x_2}^2+2x_2+1)+(4{x_3}^2+2x_3+1)+..........+(4{x_n}^2+2x_n+1)$

$=4[{x_1}^2+{x_2}^2+{x_3}^2+......+{x_n}^2]+[(2x_1+1)+(2x_2+1)+(2x_3+1)+......+(2x_n+1)]$

$=4[{x_1}^2+{x_2}^2+{x_3}^2+......+{x_n}^2]+[k_1+k_2+k_3+......+k_n]$

Thus ,we have

${k_1}^2+{k_2}^2+{k_3}^2+..........+{k_n}^2=4[{x_1}^2+{x_2}^2+{x_3}^2+......+{x_n}^2]+[k_1+k_2+k_3+......+k_n]$

Hence , when ${k_1}^2+{k_2}^2+{k_3}^2+..........+{k_n}^2$ is divided by 4, the remainder is $k_1+k_2+k_3+......+k_n$