The sequence of numbers a1, a2, a3, ..., an is defined by an

Question

The sequence of numbers  a_{1}, a_{2}, a_{3},  ... a_{n}  is defined by  a_{n} = \frac{1}{n} - \frac{1}{n+2}  for each integer n=>1 . What is the sum of the first 20 terms of the sequence?

A.  (1+\frac{1}{2})-\frac{1}{20}

B.  (1+\frac{1}{2})-(\frac{1}{21}+\frac{1}{22})

C.  1-(\frac{1}{20}+\frac{1}{22})

D.  1-\frac{1}{22}

E.  1-\frac{1}{20}-\frac{1}{22}

mvictor · Accepted Answer

this is a type of the questions that i hate the most...
i got to the solution by applying logic rather than crunching with the numbers...

fact 1. the sum will be >1. threfore, C, D, and E are out.
fact 2. since we have 1/(n+2), and last term for n is 20, then there must be smth 1/22. A is out, and B stands.

arpanpatnaik · Answer

The answer would most certainly be    . But the question needs a slight modification.  n>=1 , since the answer does consider a1 under the sum.

The sequence is :

a1 = 1-1/3
a2 = 1/2 - 1/4
a3 = 1/3 - 1/5....

We can observe that the third term in the sequence cancels the negative term in the first. A similar approach can be seen on all the terms and we would be left with 1 + 1/2 from a1 and a2 along with -1/22 and -1/21 from a20 and a19 term which could not be cancelled.

Hence the sum = (1+1/2) –  (1/21 +1/ 22). Hope it helps!

Regards,
A

Bunuel · Answer

a_1+a_2+a_3+.a_4..+a_{18}+a_{19}+a_{20}=(1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+...+(\frac{1}{18}-\frac{1}{20})+(\frac{1}{19}-\frac{1}{21})+(\frac{1}{20}-\frac{1}{22})=1+\frac{1}{2}-\frac{1}{21}-\frac{1}{22}  (all numbers except these 4 cancel).

Answer: B.

KrishnakumarKA1 · Answer

We can see that the summation will be 1-⅓ +½-¼ +⅓-⅕ + ……..1/18 -1/20 +1/19 -1/21 +1/20 - 1/22
After cancellation we get 1 +½ -1/21-1/22. Option B

BrentGMATPrepNow · Answer

Applying the formula, we get: 
a1 = (1/1  - 1/3 )
a2 = (1/2  - 1/4 )
a3 = ( 1/3  - 1/5)
a4 = ( 1/4  - 1/6)
.
.
.
.
a17 = (1/17  - 1/19 )
a18 = (1/18  - 1/20 )
a19 = ( 1/19  - 1/21)
a20 = ( 1/20  - 1/22)

So, the SUM =  (1/1  - 1/3 ) + (1/2  - 1/4 ) + ( 1/3  - 1/5) + ( 1/4  - 1/6) . . . (1/17  - 1/19 ) + (1/18  - 1/20 ) + ( 1/19  - 1/21) + ( 1/20  - 1/22)

Notice that all of the SAME COLORED fractions cancel out.
For example,  ( -1/3 ) +  1/3  = 0

The only fractions the DON'T get canceled out are those at the beginning and end of the sequence.
So, SUM =  1/1 + 1/2 - 1/21 - 1/22 
We can rewrite this as: SUM =  (1/1 + 1/2) - (1/21 + 1/22)

Answer: B

Cheers,
Brent

repellatquas · Answer

Hi, I believe in option B 1/22 will precede a minus sign and not a plus sign. Please correct that.

Bunuel · Answer

________________
Fixed. Thank you!

The sequence of numbers a1, a2, a3, ..., an is defined by an

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