practice test 2

[deepeshkumar1991]

a1,a2,a3, ......a100 upto 100 terms are given where any kth term is ak=1/k-1/k+1 for all k 1 through 100 .What is the sum of 100 terms of this sequence.
Answer is 100/101 .Can you please give the solution.

[gmbd97]

I do love this type of math.

a1= (1/1-1/2) if you put k=1
a2=(1/2-1/3)
a3=(1/3-1/4)
................
.................
.................
a98=(1/98-1/99)
a99=(1/99-1/100)
a100=(1/100-1/101)

now sum all from a1 to a100
=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)..........+(1/98-1/99)+(1/99-1/100)+(1/100-1/101)

so you can see 1/2-1/2=0, similarly all are neutralized except 1 and -1/101
therefore, we have left 1-1/101=(101-1)/101=100/101
Hope it helps!!!
GM

[tommywallach]

Hey GMB,

I'm impressed, because I can't even understand the question as it was originally written. In order to write out sequences, please make use of the word SUB to express when you're replacing subscript:

k1 --> Ksub1

I'm still confused however. You wrote: a(sub)k=1/k-1/k+1

If this were correct, then Asub1 = 1 / (1-1) / (1 + 1) = 1 / 0 / 2

But 1/0 is undefined...so I don't understand how that can be correct.

However, GMB seemed to get it just fine, so if his explanation makes sense to you, we're all good!

-t

[gmbd97]

HEY TOMMY,
yes I do agree that there is some confusion about the way question is written.
Actually 1/k-1/k+1 is meant (1/k)-(1/k+1),
If Manhattan forums would support MS word configuration, it would be helpful to write question exactly.

[tommywallach]

Hey GMB,

Unfortunately, most forums don't accept anything like Word. Either way, when writing here, please include parentheses in the necessary places. It's not too tough to do, and then everybody can understand what's happening.

Thanks! And glad you got the help you needed anyway! : )

-t