SOLUTION: 1: true or false 1000........................................998 (Sigma)7 (3/4)^n. = 63/16 (sigma)(3/4)^n N=2.........................................n=0

Algebra ->  Sequences-and-series -> SOLUTION: 1: true or false 1000........................................998 (Sigma)7 (3/4)^n. = 63/16 (sigma)(3/4)^n N=2.........................................n=0       Log On


   

Question 1131967: 1: true or false
1000........................................998
(Sigma)7 (3/4)^n. = 63/16 (sigma)(3/4)^n
N=2.........................................n=0
Found 2 solutions by Edwin McCravy, greenestamps:
Answer by Edwin McCravy(20065) About Me  (Show Source):
You can put this solution on YOUR website!



Write out the last two terms of the summation on the left side,
and write out the first two terms on the right side:



Subtract sum%28%283%2F4%29%5En%2Cn=2%2C998%29 from both sides:



matrix%281%2C3%2C%0D%0A%283%2F4%29%5E999%2B%283%2F4%29%5E1000%2C%22%22=%22%22%2C1%2B3%2F4%29

matrix%281%2C3%2C%0D%0A%283%2F4%29%5E999%2B%283%2F4%29%5E1000%2C%22%22=%22%22%2C7%2F4%29

2.6868299... × 10-125 = 1.75

That's certainly false!

Edwin

Answer by greenestamps(13215) About Me  (Show Source):
You can put this solution on YOUR website!


The other tutor appears to have overlooked part of the statement of the problem.

The two sums are the same, because each term of one sequence is the same as the corresponding term of the other sequence.

First terms:
first sequence: 7(3/4)^2 = 7(9/16) = 63/16
second sequence: (63/16)(3/4)^0 = 63/16
Second terms:
first sequence: 7(3/4)^3 = 7(27/64) = 189/64
second sequence: (63/16)(3/4)^1 = 189/64
etc., etc.

Formally....

   1000 
  (sigma) 7(3/4)^n  =
   n=2

      1000
   7*(sigma) (3/4)^n  =    [move the constant 7 outside the summation symbol]
      n=2

               1000
   (7(3/4)^2)*(sigma) (3/4)^(n-2)  =    [move 2 powers of (3/4) outside the summation symbol]
               n=2

            998
   (63/16)*(sigma) (3/4)^n    [redefine the index]
            n=0