What is the algebraic expression for sum of the sequence #7,11,15#?

2 Answers
Feb 19, 2018

#2n^2+5n#

Explanation:

The sum of the sequence means adding;

#7+11=18#

#18+15=33#

This means the sequence turns to #7,18,33#

We want to find the N'th term, we do this by finding the difference in the sequence:

#33-18=15#

#18-7=11#

Finding the difference of the differences:

#15-11=4#

To find the quadratic of the N'th term, we divide this by #2#, giving us #2n^2#

Now we take away #2n^2# from the original sequence:

#1n^2=1,4,9,16,25,36#

#therefore# #2n^2=2,8,18,50,72#

We only need the first #3# sequences:

#7-2=5#

#18-8=10#

#33-18=15#

Finding the difference between the differences:

#15-10=5#

#10-5=5#

Therefore we #+5n#

This gives us:

#2n^2+5n#

We can check this by substituting the values of #1, 2 and 3#

#2(1)^2+5(1)=2+5=7# So this works...

#2(2)^2+5(2)=8+10=18# So this works...

#2(3)^2+5(3)=18+15=33# So this works...

#therefore# the expression = #2n^2+5n#

Feb 19, 2018

Alternate...

Explanation:

The sequence is defined by: #a_n = 4n+3 #

Hence we are trying to find the sum of the first #n# terms...

#7 + 11 + 15 + ... + 4n+3 #

In sigma notation

#=> sum_(r=1) ^n 4r+3 #

We can use our knowledge of series...

#sum cn^2 + an +b -= c sum n^2 + asum n + b sum 1 #

We also know..

#sum_(r=1) ^n 1 = n #

#sum_(r=1) ^n r = 1/2 n (n+1) #

#=> sum 4n + 3 = 4sumn + 3sum1 #

#=> 4 * ( 1/2 n (n+1 ) ) + 3n #

#=> 2n(n+1) + 3n #

#=> 2n^2 +2n + 3n #

#=> 2n^2 + 5n #

#=> n(2n+5 ) #