Abhishek's Triangle

Calculus Level pending

The sum of elements in $n$ th row is given by $f\left( n \right)$ , while the sum of elements in $n$ consecutive rows starting from the first row is given by $g\left( n \right)$ .

If $s=\frac { g\left( 2013 \right) }{ f\left( 2013 \right) }$ , find the positive square root of digital sum of $s$ .

$Details$

Digital sum of $s$ is equal to sum of digits of $s$ .

Digital sum of $2013 = 2+0+1+3=6$ .

This question is part of set Serieses are fun! .

6 3 1 5 4 7 2 8

1 solution

Saharsh Rathi
Mar 10, 2017

$f\left( n \right) =\sum _{ r=1 }^{ n }{ r\left[ n-(r-1) \right] } =\frac { n(n+1)(n+2) }{ 6 }$

$g\left( n \right) =\sum _{ r=1 }^{ n }{ f\left( r \right) }$

$s=\frac { g\left( n \right) }{ f\left( n \right) } =\frac { n+3 }{ 4 }$ .

Hence, $s=\frac { 2013+3 }{ 4 } =504$

Digital sum of s $=\quad 5+0+4\quad =\quad 9$

Square root of "digital sum of s" $=\quad \sqrt { 9 } \quad =\quad 3$

Abhishek's Triangle

1 solution

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