Question

plss hurry up....need help guysssss​

Answer 1

The sum is 99.

Step-by-step explanation:

The expression is:

$[\frac{1}{2}+\frac{1}{100}]+[\frac{1}{2}+\frac{2}{100}]+...+[\frac{1}{2}+\frac{99}{100}]$

Compute the sum as follows:

$[\frac{1}{2}+\frac{1}{100}]+[\frac{1}{2}+\frac{2}{100}]+...+[\frac{1}{2}+\frac{99}{100}]=[\frac{1}{2}+\frac{1}{2}..._{(99\ times)}]+[\frac{1}{100}+\frac{2}{100}+...+\frac{99}{100}]$

                                                       $=\frac{99}{2}+\frac{1}{100}[1+2+...+99]$

The series is an AP with n = 99 terms.

The sum of n terms of an AP is:

$S=\frac{n}{2}[2a+(n-1)d]$

$[\frac{1}{2}+\frac{1}{100}]+[\frac{1}{2}+\frac{2}{100}]+...+[\frac{1}{2}+\frac{99}{100}]=[\frac{1}{2}+\frac{1}{2}..._{(99\ times)}]+[\frac{1}{100}+\frac{2}{100}+...+\frac{99}{100}]$

                                                       $=\frac{99}{2}+\frac{1}{100}[1+2+...+99]$

                                                       $=\frac{99}{2}+\frac{1}{100}[\frac{99}{2}((2\times 1)+(99-1)1]$

                                                       $=\frac{99}{2}+\frac{1}{100}[\frac{99\times 100}{2}]$

                                                       $=\frac{99}{2}+\frac{99}{2}$

                                                       $=99$

Thus, the sum is 99.

Answer 2

Step-by-step explanation:

Hey mate

Your answer is placed in the above attachment.

Hope it helps you