SOLUTION: This is so hard for me to grasp for some reason. Can someone help me, please? Convert the following partial sums into sigma notation and evaluate the sums: a) 1/(1*2^2) + 1/(2*3

Algebra ->  Sequences-and-series -> SOLUTION: This is so hard for me to grasp for some reason. Can someone help me, please? Convert the following partial sums into sigma notation and evaluate the sums: a) 1/(1*2^2) + 1/(2*3      Log On


   



Question 968173: This is so hard for me to grasp for some reason. Can someone help me, please?
Convert the following partial sums into sigma notation and evaluate the sums:
a) 1/(1*2^2) + 1/(2*3^2) + 1/(3*4^2)
b) 3 + 6 + 9 + 12 + 15
Thank you in advance for your time and assistance!

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
a) 1/(1*2^2) + 1/(2*3^2) + 1/(3*4^2)
Evaluating is just basic fraction math:

1%2F%281%2A2%5E2%29+%2B+1%2F%282%2A3%5E2%29+%2B+1%2F%283%2A4%5E2%29
1%2F%281%2A4%29+%2B+1%2F%282%2A9%29+%2B+1%2F%283%2A16%29
1%2F4+%2B+1%2F18+%2B+1%2F48
36%2F144%2B8%2F144%2B3%2F144
47%2F144

For the summation, I'll try some coloring to see if that helps:



Notice that the 1's on top stay the same.
Notice that the 2 exponents on the bottom stay the same in all terms.
Notice that the red numbers go 1,2,3
Notice that the green numbers go 2,3,4, just 1 more than the red numbers.
So if k represents a red number, then the corresponding green number is k+1.

So each term is of the form 1%2F%28red%28k%29%2Agreen%28%28k%2B1%29%29%5E2%29

And k starts at 1 and goes to 3.  We indicate that
with


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b) 3 + 6 + 9 + 12 + 15
The evaluation is simple addition, getting 45.

This one is easier than the first one.

They are the first 5 multiples of 3.  They are

3%2A1+%2B+3%2A2+%2B+3%2A3+%2B+3%2A4+%2B+3%2A5

The 3%2A%22%22 is the same on all the terms,
The numbers after the 3%2A%22%22 go 1,2,3,4,5
So if k represents the numbers on the right of the 3%2A%22%22,
then k starts at 1 and goes up to and including 5:

So each term has the form 3%2Ak

The summation is

sum%283%2Ak%5E%22%22%2Ck=1%2Cmatrix%281%2C3%2C%22%22%2C%22%22%2C5%29%29

Edwin