Question

If N=1.2 square+2.3square+3.4square+.........20.21square then find sum of digits of N.

Answer 1

Answer:

R goes to 1 to 20

N = $\sum$r(r+1)²

   =  $\sum$r(r² + 2r + 1)

 

N  =  $\sum$r³ + 2 $\sum$r² +  $\sum$r

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

    = 44100 +  5740 + 210

∴ N = 50050

Sum of digits = 10

Answer 2

Answer:

$\: \: \: \: \:$