A number theory problem by Bazmi Farooquee

Number Theory Level 1

1 + 2 + 3 + 4 + 5 + 6 + 7 + ⋯ + 999 = ? 1 + 2 + 3 + 4 + 5 + 6 +7 + \cdots + 999 = ?


The answer is 499500.

Bazmi Farooquee by Bazmi Farooquee , 20, India

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4 solutions

Bazmi Farooquee
Jul 21, 2014

By using the formula n (n + 1)/2 n = 999 ,

999 (999+1)/2 = 499500

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Using the formula for the sum of terms of an arithmetic progression , we get

S = n 2 ( a 1 + a n ) = 999 2 ( 1 + 999 ) = S=\dfrac{n}{2}(a_1+a_n)=\dfrac{999}{2}(1+999) = 499500 \color{#D61F06}\large \boxed{499500}

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Gonna Sing
Jul 23, 2014

use n/2(Un+a) 999/2(999+1) 999.500=499500

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Arun Ar
Jul 23, 2014

sum of first n numbers= n (n + 1)/2

here n = 999 ie, 999 (999+1)/2= 499500

0 Helpful 0 Interesting 0 Brilliant 0 Confused

We have 1 + 999 = 1000 2 + 998 = 1000 3 + 997 = 1000 and so on until we have 500 + 500 = 1000

Thus the sum is (500*1000)-500 = 499500 (500 is repeated)

Kamala Ramakrishnan - 6 years, 10 months ago

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