Kimberly77 wrote:
Thanks
nick1816 . could you elaborate more on this to arrive at \(\frac{n+2}{2}\) ? Thanks
Hello
Kimberly77 ,
We have discussed this question on YouTube. However, since you require assistance with just a part of the solution, I will elaborate that here.
Let’s discuss how we arrived at a\(_1\)a\(_2\)….a\(_n\) = \(\frac{(n + 2)}{2}\).
The question statement tells us that any term a\(_k\) is represented as
a\(_k\) = 1 + \(\frac{1}{k + 1}\) where k is a positive integer.
We can simplify this expression of a\(_k\) as follows:
- a\(_k\) = 1 + \(\frac{1}{(k + 1)}\)
- = \(\frac{(k + 1 + 1)}{(k + 1)}\) --- [Taking (k + 1) as the common denominator]
- = \(\frac{(k + 2)}{(k + 1)}\) ----- (1)
Now, we need to find something about the product of the first ‘n’ terms of this sequence, that is, about a1a2….an. So, let’s first find this product in its simplest form
Putting k = 1 in (1), we get a\(_1\) = \(\frac{(1 + 2)}{(1 + 1)}\) = 3/2
Putting k = 2, we get a\(_2\) = \(\frac{(2 + 2)}{(2 + 1)}\) = 4/3
Putting k = 3, we get a\(_3\) = \(\frac{(3 + 2)}{(3 + 1)}\) = 5/4
Let’s skip forward and calculate the last 2 terms:
Putting k = n – 1, we get a\(_n\)\(_-\)\(_1\) = \(\frac{(n – 1 + 2)}{(n – 1 + 1)}\) = \(\frac{(n +1)}{n}\)
Putting k = n, we get a\(_n\) = \(\frac{(n + 2)}{(n + 1)}\)
Therefore,
a\(_1\)a\(_2\)a\(_3\)….a\(_n\)\(_-\)\(_1\)a\(_n\) = \(\frac{3}{2}\) * \(\frac{4}{3}\) * \(\frac{5}{4}\) *…. * \(\frac{(n + 1)}{n}\) * \(\frac{(n + 2)}{(n + 1)}\) ---- (2)
Now closely observe (2).
The numerator (
3) of the first term (
3/
2) gets cancelled by the denominator (
3) of the second term (
4/
3).
Similarly, the numerator (
4) of second term (
4/
3) gets cancelled by the denominator (
4) of the third term (
5/
4).
The pattern would continue till the very end. The numerator (
n + 1) of a\(_n\)\(_-\)\(_1\) (
n + 1/
n) gets cancelled by the denominator (
n + 1) of an [
(n + 2)/
(n + 1)].
So, every term in the numerator and denominator gets cancelled
EXCEPT the denominator of first term a\(_1\):
2 and the numerator of the last term a\(_n\):
n + 2Thus, we finally are left with (n + 2) in the numerator and 2 in the denominator.
Therefore, we get
a\(_1\)a\(_2\)a\(_3\)….a\(_n\) = \(\frac{(n + 2)}{2}\)And done!
I hope you now understand this part and can use this to proceed further with the solution.
Best Regards,
Ashish Arora,
Quant Expert,
e-GMAT