daviesj wrote:
For all positive integers n, the sequence \(A_n\) is defined by the following relationship:
\(A_n = \frac{n-1}{n!}\)
What is the sum of all the terms in the sequence from \(A_1\) through\(A_{10}\), inclusive?
(A) \(\frac{9!+1}{10!}\)
(B) \(\frac{9(9!)}{10!}\)
(C) \(\frac{10!-1}{10!}\)
(D) \(\frac{10!}{10!+1}\)
(E) \(\frac{10(10!)}{11!}\)
One of radical theory to solve this kind of sequence problem is
Every problem which has such series, is also true for n, so this value will be true for value of n
A1 = 0, A2 = 1/2 , A3 = 1/3
Sum of A1 - A3 = 1/2 + 1/3 = 5/6
If the above sequence is correct, we can take one option after another and just substitute n = 2.
For example
\(\frac{9!+1}{10!}\) has to be true for n and nth term will be \(\frac{(n-1)!+1}{n!}\)
So we can put n = 3 and verify.
Hence the option A) 3/6 = 1/2 Eliminated.
Option B) \(\frac{9(9!)}{10!}\) - In terms of n - \(\frac{(n-1)(n-1)!}{n!}\)
2*2/6 = 2/3 Eliminated.
Option C) \(\frac{10!-1}{10!}\) - In terms of n - \(\frac{n!-1}{n!}\)
5/6 - Hold
Option D) \(\frac{10!}{10!+1}\) - In terms of n - \(\frac{n!}{n!+1}\)
6/7 - Eliminated
Option e) \(\frac{10(10!)}{11!}\)[/quote] - in terms of n - \(\frac{n(n!)}{(n+1)!}\)
Without even substituting, Eliminated !!!
Option C)
My Sir used to say, in this type of questions marking takes more time..
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