# Evaluate sum (3n + 2) for n = 0 to n = 100 .

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### 3 Answers

We have to find the sum ( 3n +2) for n = 0 to 100.

Now we can write the given expression as 3* sum ( n=0 to 100) + 2 + 100 - 0. Here we use the standard forms of summation.

The sum of the series of numbers from 0 to 100 is given by n*(n +1 )/2.

Therefore sum ( n=0 to 100) = 50*101/2

= 5050

So 3* sum ( n=0 to 100) + 2 + 100 - 0

= 3* 5050 + 2 + 100 - 0

= 15150 + 202

= 15352

**The required sum is 15352.**

To find the sum (3n+2) for n = 0 to 100.

Since the nth term a(n) = 3n+3,

the difference between the successive terms a(n+1) -a(n)= 3(n+1)+3 - (3n+3) = 3n+3+3-3n-3 = 3.

a(0) = 3, a(1) = 3+3 = 6, a(2) = 3*2+3 = 9.

consecutive terms are incresing by 3.

So the common difference is 3.

The first term a(0) = 3(0)+3 = 3.

the last term = a(100) = 3(300)+3 = 903.

The number of terms from 0 th to 100th = 101.

Therefore the sum of n terms of the AP: (1st term+last term)*(number of terms)/2 = (3+903)*(101)/2 = 453*101 = 45753.

So the sum of the 101 terms = 45753.

We'll remove the brackets:

sum (3n + 2) = sum 3n + sum 2

For n = 0 => 3*0 + 2 (1)

For n = 1 => 3*1 + 2 (2)

For n = 2 => 3*2 + 2 (3)

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

For n = 100 => 3*100 + 2 (100)

We'll add (1) + (2) + .... + (100)

(1) + (2) + .... + (100) = 3(0+1+2+...+100) + 2*101

We'll write the sum of the first 101 natural terms:

S101 = (0+100)*101/2

S101 = 101*50

S101 = 5050

(1) + (2) + .... + (100) = 3*5050 + 202

**(1) + (2) + .... + (100) = 15352**

**So, sum (3n + 2) = 15352, when n = 0 to n = 100**