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

2 Answers
Feb 11, 2017

Answer:

# sum_(r=1)^100 (r+4)/2 = 2725 #

Explanation:

We need the standard formula #sum_(r=1)^n r=1/2n(n+1)#

# :. 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 #n=100 => #

# :. sum_(r=1)^100 (r+4)/2 = 1/4(100)(109) #
# " " = 2725 #

Feb 11, 2017

Answer:

# sum_(r=1)^100 (r+4)/2 = 2725 #

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 #a=5/2# and #d=1/2# and there are #100# terms.

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 #