We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
80
3
avatar

1) \(Find \frac{1}{5} + \frac{2}{5^2} + \frac{3}{5^3} + \dotsb.\)

2) Compute the sum and enter your answer as a common fraction: \( \begin{array}{r r@{}c@{}l} & 1 &.& 11111111\ldots \\ & 0 &.& 11111111\ldots \\ & 0 &.& 01111111\ldots \\ & 0 &.& 00111111\ldots \\ & 0 &.& 00011111\ldots \\ & 0 &.& 00001111\ldots \\ & 0 &.& 00000111\ldots \\ + &&\vdots \\ \hline &&& ~~~? \end{array} \)

3) Find f(k) such that: \(\sum_{k=1}^n f(k) = n^3.\)

4) Find the sum 1^2 + 2^2 + 3^2 + ... + 1000^2.

 Jul 4, 2019
 #1
avatar+102320 
+2

Second one.....note that   

 

.11111....  =  1/9

.01111.... =  1/90

.00111.... =  1/900     etc.

 

So  we have the sum

 

1 + 1/9  + [  1/9 + 1/90 + 1/900 + 1/9000 +  1/90000 +  1/900000 +   ....... ]  =

 

1  + 1/9   +     [ (1/9)  / [ 1 - 1/10] ]  =

 

10/9    +   (1/9)  / [ 9/10 ]  =

 

10/9  +  (1/9) ( (10/9)  =

 

10/9 ( 1 + 1/9)    =

 

10/9 (10/9)  =

 

100/81

 

 

cool cool cool

 Jul 4, 2019
 #2
avatar
0

4) Sum of consecutive squares: 1/6*(2*1000 + 1) * 1000 * (1000 + 1) =333,833,500.

 

1)  sumfor(n, 1, 1000, n / 5^n) = 5 / 16

 Jul 4, 2019
edited by Guest  Jul 4, 2019
edited by Guest  Jul 4, 2019
 #3
avatar+22855 
+3

3)
Find \(f(k)\) such that: \(\sum \limits_{k=1}^n f(k) = n^3\).

 

\(\begin{array}{|rcll|} \hline \sum \limits_{k=1}^n f(k) &=& n^3 \\ \underbrace{\sum \limits_{k=1}^{n-1} f(k) }_{=(n-1)^3} + f(n) &=& n^3 \\\\ (n-1)^3 + f(n) &=& n^3 \\ f(n) &=& n^3-(n-1)^3 \\\\ \mathbf{ f(k) } &=& \mathbf{k^3-(k-1)^3} \\ \hline \end{array}\)

 

laugh

 Jul 6, 2019

2 Online Users

avatar