For all positive integers n, the sequence An is defined by

I tried a creative solution that didn't work, and was hoping you could shed light on why.

\((\frac{(A2+A10)}{2})*9\)

(since A1 is irrelevant, as it is 0, only A2-A10 are needed for the total)

Ended up with \(((\frac{1}{2!}+\frac{9}{10!})/2)*9\), which clearly is not going to lend a correct answer. What was wrong with using the formula to find the sum of a sequence, in this case?

I guess you wanted to find the sum with the formula for evenly spaced sets: (first+last)/2*(# of terms). But the given sequence is not evenly spaced, thus you cannot apply it here.

Hope it's clear.

oh my goodness, I completely overlooked that in my zeal to be clever

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..

are there more such type of questions bunnel?

Cheers Bunnels

Those questions kinda drove me crazy.

An = (n-1)/n! = n/n! - 1/n! = 1/(n-1)! - 1/n!

A1 = 1/0! - 1/1!
A2 = 1/1! - 1/2!
A3 = 1/2! - 1/3!
.
.
A10 = 1/9! - 1/10!

As we see we can cancel out all terms except 1/0! - 1/10! = 1 - 1/10! = (10! - 1 )/10!

I'm having a hard time keeping up with how you get this: \(A_1+A_2=\frac{1}{2!}=\frac{2!-1}{2!}\) and this: \(A_1+A_2+A_3=\frac{5}{3!}=\frac{3!-1}{3!}\)

Can you break it down step by step?
- Specifically, for A1+A2, why do you bring the factorial to the numerator? And what makes it 2!-1?
- For A1+A2+A3, how do you get "5" for the numerator?

1. \(A_1+A_2=\frac{1}{2!}\) but if you notice, you could write 1 as 2! - 1.

2. \(A_1+A_2+A_3=0+\frac{1}{2!}+\frac{2}{3!}=\frac{3+2}{3!}=\frac{5}{3!}=\frac{3!-1}{3!}\).

Hope it's clear.

anybody can explain why we cannot use this formula to solve this problem ?
sum of sequence term= n/2 * (last+first) ? i do it but got different answer !?

because it is factorial !!!

- 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..[/quote]

This is a really good solution!

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