If [m]a_n = 4n^2 - 1[/m] for all positive integers [m]n[/m],

Question

If  a_n = 4n^2 - 1  for all positive integers  n , what is the sum of the reciprocals of the first 5 terms of the sequence?

(A)  \frac{2}{5} 
(B)  \frac{4}{9} 
(C)  \frac{5}{11} 
(D)  \frac{1}{2} 
(E)  \frac{5}{9}

kevincan · Answer

4n^2 - 1 is a difference of squares

Anshpalsahni · Answer

Partial fraction logic applies here. Once calculating 5 numbers observe they are of the form a*b, b*c, c*d, d*e,e*f. Once reciprocated they are of the form 1/n(n+k) which based on partial fraction can be written as 1/k(1/n - 1/(n+k) for each number which would give 2 as k for all terms. Once you plug this all values except 1/2(1/a - 1/f) would cancel out.

alamin8 · Answer

So, which one should be the answer? I thought it was approximately 1/2 based on my traditional calculation, where I got 1,784/3,465. I understood later that it was not approximately 1/2. It should be less than 1/2, which is 5/11 as closest. Can you provide accurate braces to make it more straightforward?

Here is what I get,

4n^2 - 1 = (2n + 1) (2n - 1)
The reciprocal will be = 1/ 
Then, 1/2 *  
Now, first 5 terms,
1/2 *    
Cancelling the middle parts,
1/2 * (1 - 1/11)
1/2 * 10/11 
5/11

Answer: C

kevincan · Answer

Very nicely explained!

Also , if you are going to estimate 
1/3 + 1/15 + 1/35 + 1/63 + 1/99

In percents approx (33.3 + 6.6 + 3 + 1.5 +1)% =45.5%

agil31 · Answer

Please explain how you solve this question after finding out the reciprocals of the first terms without actually taking out the LCM to add these fractions. There is no proper explanation on how to arrive at the final answer while avoiding some tedious calculations

kevincan · Answer

The key is to realize that

1/(2x - 1) - 1(2x+1) = ((2x + 1)-(2x+1))/(4x^2 - 1)= 2/(4x^2 - 1)

kevincan · Answer

The key is to realize that

\frac{1}{2x-1}-\frac{1}{2x+1}=\frac{(2x+1)-(2x-1)}{(2x-1)(2x+1)}=\frac{2}{4x^2-1}

Thus, \frac{1}{4x^2-1}=\frac12\left(\frac{1}{2x-1}-\frac{1}{2x+1}\right).

For the first 5 terms, the sum is therefore \frac12\left .

Everything in the middle cancels, leaving

\frac12\left(1-\frac1{11}\right)=\frac12\left(\frac{10}{11}\right)=\boxed{\frac5{11}}.

If [m]a_n = 4n^2 - 1[/m] for all positive integers [m]n[/m],

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If [m]a_n = 4n^2 - 1[/m] for all positive integers [m]n[/m],

Prep Toolkit

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