1.1 **Problem:** Show that $$\sum_{k=0}^{n} k^{3} = \frac{n^{4}}{4} + \frac{n^{3}}{2} + \frac{n^{2}}{4}$$ **Step 1:** Recall the given formula: $$(m + 1) \sum_{k=0}^{n} k^{m} = (n + 1)^{m+1} - \sum_{r=2}^{m+1} \left[ \binom{m+1}{r} \sum_{k=0}^{n} k^{m+1-r} \right]$$ **Step 2:** Set $m=3$: $$4 \sum_{k=0}^{n} k^{3} = (n+1)^{4} - \sum_{r=2}^{4} \left[ \binom{4}{r} \sum_{k=0}^{n} k^{4-r} \right]$$ **Step 3:** Evaluate the sums inside: - For $r=2$, power is $4-2=2$, sum is $\sum_{k=0}^{n} k^{2} = \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6}$ - For $r=3$, power is $1$, sum is $\sum_{k=0}^{n} k = \frac{n^{2}}{2} + \frac{n}{2}$ - For $r=4$, power is $0$, sum is $\sum_{k=0}^{n} 1 = n+1$ **Step 4:** Substitute binomial coefficients: $$\binom{4}{2} = 6, \quad \binom{4}{3} = 4, \quad \binom{4}{4} = 1$$ **Step 5:** Write out the sum: $$4 \sum_{k=0}^{n} k^{3} = (n+1)^{4} - \left[6 \left( \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} \right) + 4 \left( \frac{n^{2}}{2} + \frac{n}{2} \right) + 1 (n+1) \right]$$ **Step 6:** Simplify inside the brackets: $$6 \left( \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} \right) = 2n^{3} + 3n^{2} + n$$ $$4 \left( \frac{n^{2}}{2} + \frac{n}{2} \right) = 2n^{2} + 2n$$ Sum inside brackets: $$2n^{3} + 3n^{2} + n + 2n^{2} + 2n + n + 1 = 2n^{3} + 5n^{2} + 4n + 1$$ **Step 7:** Expand $(n+1)^4$: $$(n+1)^4 = n^{4} + 4n^{3} + 6n^{2} + 4n + 1$$ **Step 8:** Substitute back: $$4 \sum_{k=0}^{n} k^{3} = n^{4} + 4n^{3} + 6n^{2} + 4n + 1 - (2n^{3} + 5n^{2} + 4n + 1)$$ **Step 9:** Simplify: $$4 \sum_{k=0}^{n} k^{3} = n^{4} + 2n^{3} + n^{2}$$ **Step 10:** Divide both sides by 4: $$\sum_{k=0}^{n} k^{3} = \frac{n^{4}}{4} + \frac{n^{3}}{2} + \frac{n^{2}}{4}$$ --- 1.2 **Problem:** Find $$\sum_{n=1}^{10} (2n + 3n^{3} - 5n^{2})$$ **Step 1:** Break the sum into parts: $$\sum_{n=1}^{10} 2n + \sum_{n=1}^{10} 3n^{3} - \sum_{n=1}^{10} 5n^{2}$$ **Step 2:** Use formulas: - $\sum_{n=1}^{10} n = \frac{10 \times 11}{2} = 55$ - $\sum_{n=1}^{10} n^{2} = \frac{10 \times 11 \times 21}{6} = 385$ - $\sum_{n=1}^{10} n^{3} = \left( \sum_{n=1}^{10} n \right)^{2} = 55^{2} = 3025$ **Step 3:** Calculate each sum: $$2 \times 55 = 110$$ $$3 \times 3025 = 9075$$ $$5 \times 385 = 1925$$ **Step 4:** Combine: $$110 + 9075 - 1925 = 7260$$ --- 2.1 **Problem:** Is $$\sum_{k=1}^{\infty} \frac{1}{k^{2} + 3}$$ convergent given $$\sum_{k=1}^{\infty} \left\lfloor \frac{1}{k} \right\rfloor^{2}$$ is convergent? **Step 1:** Since $\left\lfloor \frac{1}{k} \right\rfloor = 0$ for $k \geq 2$, the given series is finite and trivially convergent. **Step 2:** The series $\sum_{k=1}^{\infty} \frac{1}{k^{2} + 3}$ behaves like $\sum \frac{1}{k^{2}}$ for large $k$, which converges. **Step 3:** Therefore, yes, $\sum_{k=1}^{\infty} \frac{1}{k^{2} + 3}$ converges by comparison to a convergent p-series with $p=2$. --- 2.2 **Problem:** Does $$\sum_{k=1}^{\infty} \frac{1}{\ln(k+1)}$$ diverge given $$\sum_{k=1}^{\infty} \frac{1}{k}$$ diverges? For which $k$ does it diverge? **Step 1:** Since $\ln(k+1)$ grows slower than any power of $k$, $\frac{1}{\ln(k+1)}$ decreases slower than $\frac{1}{k^{p}}$ for any $p>0$. **Step 2:** By the integral test, $\int \frac{1}{\ln(x)} dx$ diverges, so the series diverges. **Step 3:** The series diverges for all $k \geq 1$. --- 2.3 **Problem:** Determine convergence type of $$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}}$$ **Step 1:** The absolute series is $$\sum_{n=1}^{\infty} \frac{1}{n^{2}}$$ which converges (p-series with $p=2 > 1$). **Step 2:** Since absolute convergence holds, the original series is absolutely convergent. --- 3.1 **Problem:** Determine convergence and sum of $$\sum_{n=0}^{\infty} 9^{-(n+2)} 4^{n+1}$$ **Step 1:** Rewrite terms: $$9^{-(n+2)} = \frac{1}{9^{n+2}} = \frac{1}{9^{2} 9^{n}} = \frac{1}{81} \cdot 9^{-n}$$ $$4^{n+1} = 4 \cdot 4^{n}$$ **Step 2:** Term is $$\frac{1}{81} \cdot 9^{-n} \cdot 4 \cdot 4^{n} = \frac{4}{81} \left( \frac{4}{9} \right)^{n}$$ **Step 3:** This is a geometric series with ratio $r=\frac{4}{9} < 1$, so converges. **Step 4:** Sum is $$\frac{4}{81} \cdot \frac{1}{1 - \frac{4}{9}} = \frac{4}{81} \cdot \frac{1}{\frac{5}{9}} = \frac{4}{81} \cdot \frac{9}{5} = \frac{4}{45}$$ --- 3.2 **Problem:** Determine convergence and sum of $$\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$$ **Step 1:** Use partial fractions: $$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$ **Step 2:** The series telescopes: $$\sum_{n=1}^{N} \left( \frac{1}{n} - \frac{1}{n+1} \right) = 1 - \frac{1}{N+1}$$ **Step 3:** Taking limit as $N \to \infty$: $$\lim_{N \to \infty} \left(1 - \frac{1}{N+1} \right) = 1$$ **Step 4:** Series converges and sum is 1. --- 3.3 **Problem:** Determine convergence of $$\sum_{n=1}^{\infty} \frac{(2n^{2} + 1)^{n}}{(n^{2} + 1)^{n}}$$ **Step 1:** Rewrite term: $$\left( \frac{2n^{2} + 1}{n^{2} + 1} \right)^{n}$$ **Step 2:** For large $n$, $$\frac{2n^{2} + 1}{n^{2} + 1} \approx \frac{2n^{2}}{n^{2}} = 2$$ **Step 3:** Term behaves like $2^{n}$ which grows without bound. **Step 4:** Since terms do not tend to zero, series diverges. ---