Sums I: The Gauss Sum

We are looking for the value of

$1 +2 + 3 + 4 + 5 + \ldots + 100$

Comparing it to

$1 + 2 + 3 + \ldots + n = \frac 12 n(n+1)$

We have $n = 100$ , so the answer is

$1 +2 + 3 + 4 + 5 + \ldots + 100 = \frac12 \times 100 \times 101 = 5050$

1 + 100 = 101

2 + 99 = 101

3 + 98 = 101

...

Sum(1,2,3,...98,99,100) = 101*50 = 5050

Let $S=1+2+3+4+\ldots+100$ then $\begin{aligned}2S&=1\,\,\,\,\,\,+2\,\,+3\,\,+4+\ldots+100\\ &\,\,\,\underbrace{+100+99+98+97+\ldots+1}_{\text{\$100\$ times}}\\ &=101\times 100\\ &=10100 \end{aligned}$ Therefore $S=\dfrac{10100}{2}=5050.$

The sum of an arithmetic progression is given by the formula: $s=\dfrac{n}{2}(a_1+a_n)$ where: $a_1$ = first term and $a_n$ = nth term

substituting, we get

$s=\dfrac{100}{2}(1+100)=50(101)=$ $\boxed{5050}$

1+100=101

2+99=101

3+98=101

4+97=101

....................................

.........................

........................

50+51=101

so,there is 50 times 101 and the summation is =(50x101)=5050

Use the formula

$\frac{n^2-n}{2}$

If we plug the values we get

$\frac{(100^2)-100}{2}$

Which equals to

$=5050$

Integers listed from 1 to 100 are 1,2,3,4,.............................,99,100

The common difference between these integers is 1.

Sum to list of integers is nothing but sum of numbers in Arithmetic progression

Hence, consider the formula for sum to n numbers in AP,

Sn= (n/2)[2a+(n-1)d]

a=1, d=1, n=100

Hence, the answer is 5050.

You don't have to do other thing that look at the options; they're all obviously wrong except for 5050 which is the actual answer, hahahaha.