An algebra problem by Drei Matthew

Algebra Level 2

1 + 2 + 3 + 4 + 5 + ⋯ + 99998 + 99999 + 100000 = ? 1+2+3+4+5+\cdots+99998+99999+100000 = \ ?


The answer is 5000050000.

Drei Matthew by Drei Matthew , 22, Philippines

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3 solutions

The sum of first n n natural numbers is n ( n + 1 ) 2 \dfrac{n(n+1)}{2} . Hence the answer is 100000 ( 100001 ) 2 = 5000050000 \dfrac{100000(100001)}{2} =\boxed{5000050000} .

2 Helpful 1 Interesting 1 Brilliant 0 Confused
Mahdi Raza
May 9, 2020

( n ) ( n + 1 ) 2 ⟹ ( 100000 ) ( 100001 ) 2 = 5000050000 \dfrac{(n)(n+1)}{2} \implies \dfrac{(100000)(100001)}{2} = \boxed{5000050000}

1 Helpful 1 Interesting 0 Brilliant 0 Confused
Darsh Kedia
May 13, 2020

The Sum Of First n Natural Numbers:

( n ) ( n + 1 ) 2 \Large\frac { (n)(n+1) }{ 2 }

The Answer:

( 100000 ) ( 100001 ) 2 = 5000050000 \Large\frac { (100000)(100001) }{ 2 } \quad =\quad \boxed { 5000050000 }

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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