+0  
 
0
374
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+111438 
+3

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+25481 
+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

18 Online Users

avatar
avatar