What is the product of all values generated by the expression

First change the expression to:

(n+1)((n-1)/n^2

Then do the first three:

(2+1)(2-1)/2^2
(3+1)(3-1)/3^2
(4+1)(4-1)/4^2

Notice that the plus and minus of the first and third terms cancels the denominator of the 2nd term.

Notice that for the first term the denominator will only reduce 1 power since there is no preceding term, leaving

1/2

Realize this progression will continue through the 15th term (n=16) with cancelling denominators except there is no 16th term (17-1) to cancel the 2nd power of 16 in the denominator of the 15th term.

So 1/16 remains.

Since there is no 16th term there's also no power of 17 in the denominator to cancel the (16+1) in the numerator of the 15th term.

So 17 remains.

Complete multiplication is then:

1/2*1/16*17 = 17/32

An intuitive way is to realize the question would be posed including many cancellations and that a multiple of 16 is likely to remain due to likely cancellations.

The only answer with a multiple of 16 in the denominator is 17/32.

Posted from my mobile device

Since 1 – 1/n^2 = (n^2 – 1)/n^2, the product in question is:

([2^2 – 1]/2^2)([3^2 – 1]/3^2)([4^2 – 1]/4^2)…([16^2 – 1]/16^2)

Using the difference of squares identity, we could write each difference in the numerators as the product of a sum and a difference. Notice that each sum, except for the last, can be canceled out with a corresponding factor in the denominator of the next fraction. So, we have:

([2 – 1]/2^2)([3 – 1]/3)([4 – 1]/4^2)…([16^2 – 1]/16)

Now, notice that each difference, except for the first, can be canceled out with a corresponding factor in the denominator of the preceding fraction. So, we have:

([2 – 1]/2)([16 + 1]/16) =
(1/2)(17/16) =
17/32

Answer: D