Question

Please Help
The sum of the first n terms in a sequence is always 1/n. Find the product of the first 2007 terms.

Answer 1

well, the sum of the first 2007 terms is 1/2007 right?

Answer 2

if i write \(\dfrac{1}{4}\) ,\(2\) times then

\(\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)

or if i write \(\dfrac{1}{9}\) ,\(3\) times then

\(\dfrac{1}{9}+\dfrac{1}{9}+\dfrac{1}{9}=\dfrac{1}{3}\)

or if i write \(\dfrac{1}{16}\) ,\(4\) times then

\(\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}=\dfrac{1}{4}\)

Answer 3

therefore, you are saying that the sequence might be \[\frac{ 1 }{ 2007^{2} }\times2007\]

Answer 4

so if i will write \(\Large (\dfrac{1}{n^2}) \) , \(\Large n\) times then then will be equal to

\(\Large \dfrac{1}{n}\)

Answer 5

for the sequence u are given the term \(\Large\dfrac{1}{n^2}\)

Answer 6

let me call some this seems complicated @ganeshie8 ,@freckles

Answer 7

@freckles

Answer 8

The first term, \(a_1\) must be 1
The second term is given by: \(a_1+a_2=\frac{1}{2}\), so \(a_2=\frac{1}{2}-a_1\)
The third term is given by: \(a_1+a_2+a_3=\frac{1}{3}\), therefore \(a_3=\frac{1}{3}-(a_1+a_2)=\frac{1}{3}-\frac{1}{2}\)
So, in general: \(a_i=\frac{1}{i}-\frac{1}{i-1}=-\frac{1}{i(i-1)}\)

Answer 9

The product is therefore given by:\[P_n=1\times\left(-\frac{1}{2.1}\right)\times\left(-\frac{1}{3.2}\right)\times\left(-\frac{1}{4.3}\right)...\times\left(-\frac{1}{n(n-1)}\right)\]

Answer 10

can you spot a pattern in this?

Answer 11

HINT: Look for factorials in the denominator

Answer 12

ummm..... i see a pattern but im not sure how to put it into words

Answer 13

Ignoring the signs for now, can you see this: