We are asked to find a formula for each of the given summations: a) $\sum_{i=1}^{n} (\sqrt{2i+1} - \sqrt{2i-1})$ b) $\sum_{k=1}^{100} \ln(\frac{k}{k+2})$ c) $\sum_{k=1}^{n} \frac{4}{(4k-3)(4k+1)}$ d) $\sum_{k=1}^{n} \frac{2^k+3^k}{6^k}$ e) $\sum_{k=1}^{n} \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k^2+k}}$

1. Problem Description

We are asked to find a formula for each of the given summations:

a) $\sum_{i=1}^{n} (\sqrt{2i+1} - \sqrt{2i-1})$

b) $\sum_{k=1}^{100} \ln(\frac{k}{k+2})$

c) $\sum_{k=1}^{n} \frac{4}{(4k-3)(4k+1)}$

d) $\sum_{k=1}^{n} \frac{2^k+3^k}{6^k}$

e) $\sum_{k=1}^{n} \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k^2+k}}$

2. Solution Steps

a) $\sum_{i=1}^{n} (\sqrt{2i+1} - \sqrt{2i-1})$

This is a telescoping sum.

$\sum_{i=1}^{n} (\sqrt{2i+1} - \sqrt{2i-1}) = (\sqrt{3} - \sqrt{1}) + (\sqrt{5} - \sqrt{3}) + (\sqrt{7} - \sqrt{5}) + \dots + (\sqrt{2n+1} - \sqrt{2n-1})$

The terms cancel out, leaving us with $\sqrt{2n+1} - \sqrt{1} = \sqrt{2n+1} - 1$.

b) $\sum_{k=1}^{100} \ln(\frac{k}{k+2})$

Using the logarithm property $\ln(\frac{a}{b}) = \ln(a) - \ln(b)$, we have

$\sum_{k=1}^{100} \ln(\frac{k}{k+2}) = \sum_{k=1}^{100} (\ln(k) - \ln(k+2))$

This is also a telescoping sum.

$= (\ln(1) - \ln(3)) + (\ln(2) - \ln(4)) + (\ln(3) - \ln(5)) + \dots + (\ln(99) - \ln(101)) + (\ln(100) - \ln(102))$

$= \ln(1) + \ln(2) - \ln(101) - \ln(102) = \ln(1) + \ln(2) - \ln(101 \cdot 102)$

Since $\ln(1) = 0$, the sum is $\ln(2) - \ln(10302) = \ln(\frac{2}{10302}) = \ln(\frac{1}{5151})$

c) $\sum_{k=1}^{n} \frac{4}{(4k-3)(4k+1)}$

We can decompose the fraction into partial fractions.

$\frac{4}{(4k-3)(4k+1)} = \frac{A}{4k-3} + \frac{B}{4k+1}$

$4 = A(4k+1) + B(4k-3)$

Let $k = \frac{3}{4}$. Then $4 = A(3+1) + B(0) \implies 4 = 4A \implies A = 1$

Let $k = -\frac{1}{4}$. Then $4 = A(0) + B(-1-3) \implies 4 = -4B \implies B = -1$

So, $\frac{4}{(4k-3)(4k+1)} = \frac{1}{4k-3} - \frac{1}{4k+1}$

$\sum_{k=1}^{n} \frac{4}{(4k-3)(4k+1)} = \sum_{k=1}^{n} (\frac{1}{4k-3} - \frac{1}{4k+1})$

This is a telescoping sum.

$= (\frac{1}{1} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{9}) + (\frac{1}{9} - \frac{1}{13}) + \dots + (\frac{1}{4n-3} - \frac{1}{4n+1}) = 1 - \frac{1}{4n+1} = \frac{4n+1-1}{4n+1} = \frac{4n}{4n+1}$

d) $\sum_{k=1}^{n} \frac{2^k+3^k}{6^k} = \sum_{k=1}^{n} (\frac{2^k}{6^k} + \frac{3^k}{6^k}) = \sum_{k=1}^{n} (\frac{1}{3^k} + \frac{1}{2^k}) = \sum_{k=1}^{n} (\frac{1}{3})^k + \sum_{k=1}^{n} (\frac{1}{2})^k$

Using the formula for the sum of a geometric series $\sum_{k=1}^{n} ar^{k-1} = a\frac{1-r^n}{1-r}$, we have $\sum_{k=1}^{n} r^k = \frac{r(1-r^n)}{1-r}$.

$\sum_{k=1}^{n} (\frac{1}{3})^k = \frac{\frac{1}{3}(1-(\frac{1}{3})^n)}{1-\frac{1}{3}} = \frac{\frac{1}{3}(1-\frac{1}{3^n})}{\frac{2}{3}} = \frac{1}{2}(1-\frac{1}{3^n})$

$\sum_{k=1}^{n} (\frac{1}{2})^k = \frac{\frac{1}{2}(1-(\frac{1}{2})^n)}{1-\frac{1}{2}} = \frac{\frac{1}{2}(1-\frac{1}{2^n})}{\frac{1}{2}} = 1-\frac{1}{2^n}$

$\sum_{k=1}^{n} \frac{2^k+3^k}{6^k} = \frac{1}{2}(1-\frac{1}{3^n}) + 1 - \frac{1}{2^n} = \frac{1}{2} - \frac{1}{2 \cdot 3^n} + 1 - \frac{1}{2^n} = \frac{3}{2} - \frac{1}{2 \cdot 3^n} - \frac{1}{2^n}$

e) $\sum_{k=1}^{n} \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k^2+k}} = \sum_{k=1}^{n} \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k}\sqrt{k+1}} = \sum_{k=1}^{n} (\frac{\sqrt{k+1}}{\sqrt{k}\sqrt{k+1}} - \frac{\sqrt{k}}{\sqrt{k}\sqrt{k+1}}) = \sum_{k=1}^{n} (\frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k+1}})$

This is a telescoping sum.

$= (\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}) + \dots + (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}) = \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{n+1}} = 1 - \frac{1}{\sqrt{n+1}}$

3. Final Answer

a) $\sqrt{2n+1} - 1$

b) $\ln(\frac{1}{5151})$

c) $\frac{4n}{4n+1}$

d) $\frac{3}{2} - \frac{1}{2 \cdot 3^n} - \frac{1}{2^n}$

e) $1 - \frac{1}{\sqrt{n+1}}$