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. We can write out the first few terms:

$(\sqrt{3}-\sqrt{1}) + (\sqrt{5}-\sqrt{3}) + (\sqrt{7}-\sqrt{5}) + ... + (\sqrt{2n+1} - \sqrt{2n-1})$

The intermediate 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 properties of logarithms, we can write this as

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

This is also a telescoping sum. Writing out the first few terms:

$(\ln(1) - \ln(3)) + (\ln(2) - \ln(4)) + (\ln(3) - \ln(5)) + (\ln(4) - \ln(6)) + ... + (\ln(99) - \ln(101)) + (\ln(100) - \ln(102))$

We have $\ln(1) + \ln(2) - \ln(101) - \ln(102) = \ln(1) + \ln(2) - \ln(101) - \ln(102) = \ln(1) + \ln(2) - \ln(101 \cdot 102) = \ln(\frac{2}{10302}) = \ln(\frac{1}{5151})$

Therefore the sum is $\ln(1) + \ln(2) - \ln(101) - \ln(102) = \ln(1*2) - \ln(101*102) = \ln(2) - \ln(10302) = \ln(\frac{2}{10302}) = \ln(\frac{1}{5151}) = -\ln(5151)$

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

We can decompose the fraction using 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 $4 = A(4(\frac{3}{4}) + 1) + B(0) \implies 4 = 4A \implies A=1$

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

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

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

This is a telescoping sum. Let's write out the first few terms:

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

The terms cancel out, leaving $1 - \frac{1}{4n+1} = \frac{4n+1-1}{4n+1} = \frac{4n}{4n+1}$

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

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

These are geometric series. $\sum_{k=1}^{n} r^k = \frac{r(1-r^n)}{1-r}$

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

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

Therefore, the sum is $\frac{1}{2}(1 - (\frac{1}{3})^n) + 1 - (\frac{1}{2})^n = \frac{3}{2} - \frac{1}{2}(\frac{1}{3})^n - (\frac{1}{2})^n = \frac{3}{2} - \frac{1}{2 \cdot 3^n} - \frac{1}{2^n}$

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.

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

The intermediate terms cancel out, leaving us with

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

3. Final Answer

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

b) $-\ln(5151)$

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

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

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