# How do you find the partial sum of #Sigma (n+4)/2# from n=1 to 100?

##### 2 Answers

#### Answer:

#### Explanation:

We need the standard formula

# :. sum_(k=1)^n (r+4)/2 = 1/2 sum_(k=1)^n (r+4) #

# " "= 1/2 {sum_(k=1)^n r +sum_(k=1)^n 4}#

# " "= 1/2 {1/2n(n+1) +4n}#

# " "= 1/2 *1/2{n(n+1) +8n}#

# " "= 1/4{n^2+n +8n}#

# " "= 1/4{n^2+9n}#

# " "= 1/4n(n+9)#

Put

# :. sum_(r=1)^100 (r+4)/2 = 1/4(100)(109) #

# " " = 2725 #

#### Answer:

#### Explanation:

Another approach using the AP formula:

Write out the first few terms to establish the pattern;

# sum_(r=1)^100 (r+4)/2 = 5/2 + 6/2 + 7/2 + ... 104/2 #

The terms for an Arithmetic Progression (AP) with

Using the standard AP formula:

# S_n = 1/2n{2a + (n-1)d} #

We get;

# S_100 = (100)/2{2(5/2) + (99)(1/2)} #

# " " = 50{5 + 99/2} #

# " " = 2725 #