Adding 100 different number

Question

How do I compute

$1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50+51+52+53+54+55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70+71+72+73+74+75+76+77+78+79+80+81+82+83+84+85+86+87+88+89+90+91+92+93+94+95+96+97+98+99+100$

in an efficient way without using calculator? So far I only add up to $19$ and I get $171$ can anyone tell me how to do this faster? Thanks.

Answer 1

Notice that you can add $1+99$ to get $100$. You can also add $2+98$ to get $100$. You can do this all the way up to $49+51$, having only left out $50$ and $100$. This means the total is $$ 49(100)+100+50 $$ This is a classic problem with a fun story behind it: http://mathcentral.uregina.ca/QQ/database/QQ.02.06/jo1.html

More generally this type of sum is an arithmetic series and can be calculated for $n$ terms from $a_1$ to $a_n$ as $$ \frac{1}{2}n(a_1+a_n) $$

Answer 2

By pairing the first number with the last, you get $101$. Pairing the second with the second to last also gives $101$. Continuing in this fashion, you have $\frac{100}{2}$ pairs of $101$, so you have $101\cdot\frac{100}{2}$ total.

In general, the sum from $1$ to $n$ is $\frac{n(n+1)}{2}$.

Answer 3

That's an induction problem. (Are you familiar with that?)

Afirmation: $1 + 2 + \dots + n = \frac{n(n+1)}{2}$

In you case, $n = 100$, so, the sum will be $5050$.