An algebra problem by pranay reddy chintu

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

a simple formula=n*(n+1)/2

20*(21)/2

420/2

210

1+2+3+4................+20

=1+20+2+19+3+18.................

=21+21+21...............(10 times)

= $\boxed{210}$

$1 + 20 = 21$ $2 + 19 = 21$ $3 + 18 = 21$ $4 + 17 = 21$ $\cdots$ .

Total of $10$ $21$ exists.

So, $10 \times 21 = \boxed { 210 }$ .

Actually there is a pattern here,just consider that 21 is n since 20+1=21,19+2=21,thus use this formula n(n-1)/2=21(20)/2=420/2=210

1+2+3+....20 The formula(un) is n+1 Sum of arithmetics number(sn) is n/2 (1st number x un) Then S20 = 20/2 (1(20+1) So the answer is 210

Use the Gauss's sum

n*(n+1)/2

=20*(20+1)/2

=210