Question

The number of positive integers n in the set $\{1,2,3....100\}$ for which the number $\frac{1^2+2^2+3^2+.....n^2}{1+2+3+.....n}$ is an integer is:

• A. $33$
• B. $34$
• C. $50$
• D. $100$

Answer 1

The correct option is B $34$

Given:

$=\frac{1^2+2^2+3^2+.....n^2}{1+2+3+.....n}$

$=\frac{2n(n+1)(2n+1)}{6n(n+1)}$ (Formula of summation)

$=\frac{2n+1}{3}$

For $n=1,\epsilon I^{+}$

For $n=4,\epsilon I^{+}$

For $n=7,\epsilon I^{+}$

$\therefore$ No. of positive integers $=\frac{100-1}{3}+1=34$ $[\because T_n=a+(n-1\left)d\right]$