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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

Respuesta :

jakethecat
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:

Sum to n terms of an Airthmetic Progression is given by

    [tex]S_{n}=\frac{n \times (n+1)}{2} \text{or}\\\\ S_{n}=(a+l)\times \frac{n}{2}[/tex]

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.

[tex]S_{n}=\frac{100}{2}\times (1+100)\\\\S_{n}=50 \times 101\\\\S_{n}=5050[/tex]

Option D

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