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.
+128408 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
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
+26367 3)
Find \(f(k)\) such that: \(\sum \limits_{k=1}^n f(k) = n^3\).
\(\begin{array}{|rcll|} \hline \sum \limits_{k=1}^n f(k) &=& n^3 \\ \underbrace{\sum \limits_{k=1}^{n-1} f(k) }_{=(n-1)^3} + f(n) &=& n^3 \\\\ (n-1)^3 + f(n) &=& n^3 \\ f(n) &=& n^3-(n-1)^3 \\\\ \mathbf{ f(k) } &=& \mathbf{k^3-(k-1)^3} \\ \hline \end{array}\)
