Question

To find the sum of consecutive integers, starting at 1, we use the formula , where n is the last number. Find the sum of the numbers from 1 to 100. 1 + 2 + 3 + ... + 100 A. 101 B. 1,001 C. 5,000 D. 5,050 Please select the best answer from the choices provided

Answer 1

1+2+3+4+...+n=n(n+1)/2
1+2+3+4+...+100=100(101)/2
=101(50)
=5050(D)

Hope it helps!!!

Answer 2

Sum to n terms of an Airthmetic Progression is given by

`S_(n)=(n * (n+1))/(2) \text{or}\\\\ S_(n)=(a+l)* (n)/(2)`

a=First term

l=Last term

Now, the given series is

1+2+3+4+5+6+.......+100

It is an Aritmetic progression , as difference between two consecutive terms is same.

`S_(n)=(100)/(2)* (1+100)\\\\S_(n)=50 * 101\\\\S_(n)=5050`

Option D