I'm guessing the sum is supposed to be
![\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5E%5Cinfty%5Cfrac%7B10%7D%7B%285k-1%29%285k%2B4%29%7D)
Split the summand into partial fractions:
![\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%285k-1%29%285k%2B4%29%7D%3D%5Cdfrac%20a%7B5k-1%7D%2B%5Cdfrac%20b%7B5k%2B4%7D)
![1=a(5k+4)+b(5k-1)](https://tex.z-dn.net/?f=1%3Da%285k%2B4%29%2Bb%285k-1%29)
If
, then
![1=b(-4-1)\implies b=-\frac15](https://tex.z-dn.net/?f=1%3Db%28-4-1%29%5Cimplies%20b%3D-%5Cfrac15)
If
, then
![1=a(1+4)\implies a=\frac15](https://tex.z-dn.net/?f=1%3Da%281%2B4%29%5Cimplies%20a%3D%5Cfrac15)
This means
![\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}](https://tex.z-dn.net/?f=%5Cdfrac%7B10%7D%7B%285k-1%29%285k%2B4%29%7D%3D%5Cdfrac2%7B5k-1%7D-%5Cdfrac2%7B5k%2B4%7D)
Consider the
th partial sum of the series:
![S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)](https://tex.z-dn.net/?f=S_n%3D2%5Cleft%28%5Cdfrac14-%5Cdfrac19%5Cright%29%2B2%5Cleft%28%5Cdfrac19-%5Cdfrac1%7B14%7D%5Cright%29%2B2%5Cleft%28%5Cdfrac1%7B14%7D-%5Cdfrac1%7B19%7D%5Cright%29%2B%5Ccdots%2B2%5Cleft%28%5Cdfrac1%7B5n-1%7D-%5Cdfrac1%7B5n%2B4%7D%5Cright%29)
The sum telescopes so that
![S_n=\dfrac2{14}-\dfrac2{5n+4}](https://tex.z-dn.net/?f=S_n%3D%5Cdfrac2%7B14%7D-%5Cdfrac2%7B5n%2B4%7D)
and as
, the second term vanishes and leaves us with
![\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5E%5Cinfty%5Cfrac%7B10%7D%7B%285k-1%29%285k%2B4%29%7D%3D%5Clim_%7Bn%5Cto%5Cinfty%7DS_n%3D%5Cfrac17)
Answer:
an octagon has 3 more sides then a pentagon
Step-by-step explanation:
A pentagon has 5 sides while a octagon has 8 sides and 8 -5=3
We are asked to express r in terms of A, P, and t.
We first divide both sides of the equation by t, which gives us
![\displaystyle{ \frac{A}{t}=P(1+r)](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%20%5Cfrac%7BA%7D%7Bt%7D%3DP%281%2Br%29%20)
,
then, dividing both sides by P, we have
![\displaystyle{ \frac{A}{Pt}=1+r](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%20%5Cfrac%7BA%7D%7BPt%7D%3D1%2Br%20)
.
Swap the sides:
Finally subtracting 1 from both sides gives us
![\displaystyle{ r=\frac{A}{Pt}-1](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%20r%3D%5Cfrac%7BA%7D%7BPt%7D-1)
.
Answer:
sin33°=y/11
Solve for y
Step-by-step explanation:
Answer:
c) -x^3 + x^2 - 1
Step-by-step explanation:
Given: u (x) = x^5 - x^4 +x^2 and v(x) = -x^2
(u/v)(x) = u(x)/v(x)
Now plug in the given functions in the above formula, we get
= (x^5 - x^4 + x^2) / -x^2
We can factorize the numerator.
In x^5 - x^4 + x^2. the common factor is x^2, so we can take it out and write the remaining terms in the parenthesis.
= x^2 (x^3 - x^2 + 1) / - x^2
Now we gave x^2 both in the numerator and in the denominator, we can cancel it out.
(u/v)(x) = (x^3 - x^2 + 1) / -1
When we dividing the numerator by -1, we get
(u/v)(x) = -x^3 + x^2 - 1
Answer: c) -x^3 + x^2 - 1
Hope you will understand the concept.
Thank you.