In the sequence a1, a2, a3, ..., a100 the kth term is defined by ak

Question

In the sequence a1, a2, a3, ..., a100 the kth term is defined by  a_k = \frac{1}{k} - \frac{1}{k+1}  for all integers k from 1 through 100. What is the sum of the 100 terms of this sequence?

A)  \frac{1}{10,100}

B)  \frac{1}{101}

C)  \frac{1}{100}

D)  \frac{100}{101}

E)  1

k1.saurabh · Answer

It is given that ak=1/k−1/k+1

a1= 1/1-1/2

a2= 1/2-1/3

a3= 1/3-1/4 
....
..
.
a99=1/99-1/100

a100= 1/100-1/101

When we add all the terms, it can be seen that all the terms from 1/2 to 1/100 will cancel each other because of opposite signs. Only remaining terms will be 1/1 and -1/101.

So sum = 1/1-1/101= 101-1/101= 100/101

Answer is D

ScottTargetTestPrep · Answer

Starting with the first 4 terms we have:

a1 + a2 + a3 + a4 = (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5)

We see that all terms cancel except for the first term and the final term. Combining those two terms, we have 1 - 1/5 = 4/5.

Extending this pattern to all 100 numbers, we have:

1 - 1/101 = 101/101 - 1/101 = 100/101

Answer: D

NandishSS · Answer

a_k= \frac{1}{k} - \frac{1}{k+1}

a_1= \frac{1}{1} - \frac{1}{1+1}

a_2= \frac{1}{2} - \frac{1}{2+1}

a_100= \frac{1}{100} - \frac{1}{100+1}

Sum = (1/1) - (1/2) + (1/2) - (1/3) ...... (1/100+1/101 )

Sum = 1- (1/101)

Sum = 100/101

Hence D

byebye1220 · Answer

ScottTargetTestPrep  why does using the formual for the sum of the terms of an arithmetic sequence not work here?

Bunuel · Answer

Because the terms are NOT in arithmetic progression:

a_1 = 1 - \frac{1}{2} = \frac{1}{2} ,

a_2 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} ,

a_3 = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} ,

a_4 = \frac{1}{4} - \frac{1}{5} = \frac{1}{20} ,

...

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In the sequence a1, a2, a3, ..., a100 the kth term is defined by ak

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In the sequence a1, a2, a3, ..., a100 the kth term is defined by ak

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