Sum of large numbers

Sum of ( 1 + 2 +3 + 4 + 5 + . . . . . . . . . . + 10^100 ) = ( 10^100 / 2 ) + ( 10^200 / 2 ) = 5 . 10^99 + 5 . 10^199 = b : So : Sum of All the digits of ( b ) equal ( 5 + 5 = 10 ) .

$\begin{aligned} S & = 1 + 2 + 3 + \cdots + 10^{100} & \small \blue{\text{Note that }1+2+3+\cdots + n= \frac {n(n+1)}2} \\ & = \frac {10^{100}(10^{100}+1)}2 \\ & = 5 \times 10^{99}(10^{100}+1) \\ & = 5\underbrace{0000 \cdots 0000}_{\text{number of 0s}=99}(1\underbrace{0000 \cdots 0000}_{\text{number of 0s}=99}1) \\ & = 5\underbrace{0000 \cdots 0000}_{\text{number of 0s}=99}5\underbrace{0000 \cdots 0000}_{\text{number of 0s}=99} \end{aligned}$

Therefore the sum of digits of $S$ is $5+5 = \boxed {10}$ .

step by step solution