Question

Q.10. Prove that on dividing sum of square
of two consecutive numbers by 4, the remain-
der is always 1.​

Answer 1

Answer:

remainder should be 2

question is incorrect

Answer 2

Answer:here's the proof.

Explanation:let n be a number.

reqd sum:

${n}^{2} + {(n + 1)}^{2}$

$= {n}^{2} + {n + 1 + 2n}^{2}$

$= 2 {n}^{2} + 2n + 1$

dividing it with 4

$(2( {n}^{2} + n) + 1) + \div 4$

so remainder is 1

since

$2( {n}^{2} + n )$

is always divisible by 4

please mark as the brainliest answer