Arithmetic progression 1 to 100

We have , $S = 1+2+3+4 ...... +100 \rightarrow(i)$

If we reverse the series the summation still remains the same , $S= 100 + 99 +98+.......+2+1 \rightarrow(ii)$

adding $i$ and $ii$ we have , $2S = 101 + 101 +.......+ 101$

Now $101 +101 +101 +...... +101$ goes on for $100$ terms so , $2S = 100 \times 101$ $S = \frac{100\times 101}{2} \Rightarrow \boxed{5050}$

This is how Gauss figured it out

From the above method we have the general formula for finding the summation of $n$ consecutive natural numbers, $\sum_{i =1}^{n} k = \frac{n\times(n+1)}{2}$

Gauss calculated the sum of 1 and 100, 99 and 2, 98 and 3, 97 and 4 and so forth then he multiply it by 50 which yields to 101 times 50

0,1,2,3,,,,,,,99,100 are 101 Numbers and each 2 numbers of them = 100

so 101/2 =50.5

50.5 * 100 = 5050

Thanks!!

Sn=n(an+a1)/2

n=100, an=100, a1=1

100(100+1)/2=50(101)=5050

Your question is correct but the answer should be 4949 Coz in the question it has mentioned the sum of all terms BETWEEN 1 and 100..though I got it right but you should edit,So others do not have any problem....

Since we know that the first term is 1 and the last term is 100 we can use these values to calculate the sum of all terms. Therefore 100 is multiplied by 100 +1 divided by 2. This provides us with the solution of 5050.