An algebra problem by Devamsh Kondle

$1+2+3+\cdots+998+999+1000$ $=\frac{1000\times 1001}{2}$ $=\boxed{500500}.$

multiply the last number and the number succeeding it, and then divide the product by two................!

1+2+3+4...............+1000

=1+1000+2+999+3+998........................

=1001+1001+1001.................(500 Times)

= $\boxed{500500}$

The answer is $\dfrac{1000 × 1001}{2} = 500 × 1001 = \color{#69047E}{\boxed{500500}}$ .

Use the given formula n(n-1),consider that n is 1001,since the pattern is 1000+1 so equal to 1001(1001-1)=1001(1000)=1001000/2=500500

Using ROOT

void j(){ int s = 0; for (int i=1;i<1001;i++){ s = s+i; } cout<<s<<'\n'; }

1+1000=1001, 2+999=1001 until 500+501=1001 1001 times 500 is 500500

S(1000)=(1000/2) (1000+1)=500 1001=5005*100=500500