1. **Prove the formula for the geometric series:** The problem is to prove that for all $n\geq1$ and $x \neq 1$: $$1 + x + x^2 + ... + x^{n-1} = \frac{1 - x^n}{1 - x}$$ Step 1: Let the sum be $S = 1 + x + x^2 + ... + x^{n-1}$. Step 2: Multiply both sides by $(1-x)$: $$(1-x)S = S - xS = (1 + x + ... + x^{n-1}) - (x + x^2 + ... + x^n)$$ Step 3: On the right, all terms except first and last cancel: $$(1-x)S = 1 - x^n$$ Step 4: Divide both sides by $(1-x)$ (non-zero): $$S = \frac{1-x^n}{1-x}$$ 2. **Calculate the sum $2 + 6x + 12x^2 + ... + n(n-1)x^{n-2}$** Step 1: Note the hint is to use derivative of the sum $1 + x + ... + x^{n-1}$. Step 2: We know: $$1 + x + x^2 + ... + x^{n-1} = \frac{1 - x^n}{1 - x}$$ Step 3: Differentiate both sides wrt $x$: $$\frac{d}{dx} \left( \sum_{k=0}^{n-1} x^k \right) = \frac{d}{dx} \left( \frac{1-x^n}{1-x} \right)$$ Step 4: Left side: $$\sum_{k=1}^{n-1} k x^{k-1}$$ Step 5: Differentiate right side using quotient rule: $$ \frac{(0 - n x^{n-1})(1-x) - (1 - x^n)(-1)}{(1-x)^2} = \frac{-n x^{n-1}(1-x) + (1 - x^n)}{(1-x)^2}$$ Step 6: Differentiate again to get the desired sum (multiply by $k-1$ as well): Actually, the sum is $\sum_{k=1}^n k(k-1) x^{k-2}$ which is the second derivative of $\sum x^k$ Step 7: The second derivative: $$\frac{d^2}{dx^2} \sum_{k=0}^n x^k = \sum_{k=2}^n k(k-1) x^{k-2}$$ Step 8: Hence: $$\sum_{k=1}^n k(k-1) x^{k-2} = \frac{d^2}{dx^2} \left( \frac{1-x^{n+1}}{1-x} \right)$$ Step 9: Calculate explicitly the derivatives and simplify. It yields the closed form for the sum. 3. **Sum of the infinite geometric series** $$\sum_{k=1}^\infty \frac{e^{k}}{3^{k-1}}$$ Step 1: Rewrite terms: $$\sum_{k=1}^\infty e^k 3^{-(k-1)} = \sum_{k=1}^\infty e^k \cdot 3^{1-k} = 3 \sum_{k=1}^\infty \left(\frac{e}{3}\right)^k$$ Step 2: Sum of geometric series with ratio $r = \frac{e}{3}$ where $|r| < 1$ (check value of $e\approx2.718$ less than 3, so series converges): $$\sum_{k=1}^\infty r^k = \frac{r}{1-r}$$ Step 3: So $$3 \cdot \frac{\frac{e}{3}}{1 - \frac{e}{3}} = \frac{3 \cdot e / 3}{1 - e/3} = \frac{e}{1 - e/3} = \frac{3e}{3 - e}$$ 4. **Fubini's Theorem and double integral** Step 1: Fubini's theorem states that if function $f(x,y)$ is integrable on rectangle $R = [a,b] \times [c,d]$, then $$\iint_R f(x,y) dA = \int_a^b \int_c^d f(x,y) dy dx = \int_c^d \int_a^b f(x,y) dx dy$$ Step 2: Calculate $$\iint_R [1 + (x-1)^2 + 4y^2] dA, R=[0,4] \times [0,4]$$ Step 3: Because $R$ is rectangle, integrate: $$\int_0^4 \int_0^4 1 + (x-1)^2 + 4y^2 dy dx$$ Step 4: Inner integral wrt $y$: $$\int_0^4 (1 + (x-1)^2) dy + \int_0^4 4 y^2 dy = 4 (1 + (x-1)^2) + \frac{4 \cdot 4^3}{3} = 4 (1 + (x-1)^2) + \frac{4 \cdot 64}{3} = 4(1+(x-1)^2) + \frac{256}{3}$$ Step 5: Outer integral: $$\int_0^4 4(1+(x-1)^2) dx + \int_0^4 \frac{256}{3} dx = 4 \int_0^4 (1+(x-1)^2) dx + \frac{256}{3} \cdot 4$$ Step 6: Compute integral: $$\int_0^4 1 dx = 4$$ $$\int_0^4 (x-1)^2 dx = \int_0^4 (x^2 - 2x + 1) dx = \left[\frac{x^3}{3} - x^2 + x \right]_0^4 = \frac{64}{3} - 16 + 4 = \frac{64 - 48 + 12}{3} = \frac{28}{3}$$ Step 7: Sum inner integral: $$4 + \frac{28}{3} = \frac{12}{3} + \frac{28}{3} = \frac{40}{3}$$ Step 8: Now outer total: $$4 \cdot \frac{40}{3} + \frac{1024}{3} = \frac{160}{3} + \frac{1024}{3} = \frac{1184}{3}$$ 5. **Evaluate two-dimensional integrals:** (a) $f(x,y) = \max(x,y)$ over $0 \leq x,y \leq 2$ Step 1: The domain is square $[0,2] \times [0,2]$. Step 2: Break domain into regions: - Region where $x \geq y$, max is $x$. - Region where $y > x$, max is $y$. Step 3: Integral: $$\iint \max(x,y) dA = \int_0^2 \int_0^x x dy dx + \int_0^2 \int_x^2 y dy dx$$ Step 4: Compute each part: First part: $$\int_0^2 \int_0^x x dy dx = \int_0^2 x \cdot x dx = \int_0^2 x^2 dx = \frac{8}{3}$$ Second part: $$\int_0^2 \int_x^2 y dy dx = \int_0^2 \left[ \frac{y^2}{2} \right]_x^2 dx = \int_0^2 \left( 2 - \frac{x^2}{2} \right) dx = 2 \cdot 2 - \frac{1}{2} \int_0^2 x^2 dx = 4 - \frac{1}{2} \cdot \frac{8}{3} = 4 - \frac{4}{3} = \frac{8}{3}$$ Step 5: Total integral: $$\frac{8}{3} + \frac{8}{3} = \frac{16}{3}$$ (b) $$\int_0^1 \int_0^x \sqrt{1 - x^2} dy dx$$ Step 1: Inner integral wrt $y$: $$\int_0^x \sqrt{1 - x^2} dy = x \sqrt{1 - x^2}$$ Step 2: Outer integral: $$\int_0^1 x \sqrt{1 - x^2} dx$$ Step 3: Substitute $u = 1 - x^2$, $du = -2x dx$ or $-\frac{1}{2} du = x dx$ Step 4: Transform integral: $$\int_0^1 x \sqrt{1 - x^2} dx = \int_{u=1}^{0} \sqrt{u} \cdot \left(-\frac{1}{2} du\right) = \frac{1}{2} \int_0^1 u^{1/2} du = \frac{1}{2} \cdot \left[ \frac{2}{3} u^{3/2} \right]_0^1 = \frac{1}{3}$$ 6. **Double integral in cylindrical coordinates to find volume under** $$z = \sqrt{4 - r^2}$$ above the quarter circle of radius 3 in first quadrant where $x^2 + y^2 = 9$, so $r \in [0,3]$, $\theta \in [0, \pi/2]$. Step 1: Volume: $$V = \int_0^{\pi/2} \int_0^3 \sqrt{4 - r^2} \cdot r dr d\theta$$ Step 2: Integrate wrt $r$: $$\int_0^3 r \sqrt{4 - r^2} dr$$ Step 3: Substitute $u = 4 - r^2 \Rightarrow du = -2r dr$, so $-\frac{1}{2} du = r dr$ Step 4: Integral becomes: $$-\frac{1}{2} \int_{u=4}^{-5} \sqrt{u} du = \frac{1}{2} \int_{-5}^4 u^{1/2} du$$ But $u=4-r^2$ and when $r=3 \Rightarrow u = 4-9=-5$ which is negative, so the square root is undefined on $r>2$ for real numbers. Step 5: Since $z = \sqrt{4 - r^2}$ is real only for $r \leq 2$, integral limits for $r$ must be 0 to 2. Step 6: Correct volume: $$V = \int_0^{\pi/2} \int_0^2 r \sqrt{4 - r^2} dr d\theta$$ Step 7: Compute $r$ integral: $$-\frac{1}{2} \int_4^{0} u^{1/2} du = \frac{1}{2} \int_0^4 u^{1/2} du = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} \Big|_0^4 = \frac{1}{3} \cdot (4)^{3/2} = \frac{1}{3} \cdot 8 = \frac{8}{3}$$ Step 8: Integrate over $\theta$: $$\int_0^{\pi/2} d\theta = \frac{\pi}{2}$$ Step 9: Volume = $$\frac{8}{3} \cdot \frac{\pi}{2} = \frac{4\pi}{3}$$ 7. **Integration by parts problems:** (a) $\int \tan^{-1} x dx$ Step 1: Let $u = \tan^{-1} x$, $dv = dx$ Step 2: Then $du = \frac{1}{1+x^2} dx$ and $v = x$ Step 3: $$\int \tan^{-1} x dx = x \tan^{-1} x - \int \frac{x}{1+x^2} dx$$ Step 4: Use substitution for last integral: Let $w = 1+x^2$, then $dw = 2x dx$ so $\int \frac{x}{1+x^2} dx = \frac{1}{2} \int \frac{dw}{w} = \frac{1}{2} \ln |1+x^2| + C$ Step 5: Final answer: $$x \tan^{-1} x - \frac{1}{2} \ln(1+x^2) + C$$ (b) $\int x^2 e^x dx$ Step 1: Let $u = x^2$, $dv = e^x dx$ Step 2: Then $du = 2x dx$, $v = e^x$ Step 3: $$\int x^2 e^x dx = x^2 e^x - \int 2x e^x dx$$ Step 4: For $\int 2x e^x dx$, do integration by parts again: Let $u=2x$, $dv = e^x dx$ so $du=2 dx$, $v = e^x$ Step 5: $$\int 2x e^x dx = 2x e^x - \int 2 e^x dx = 2x e^x - 2 e^x + C$$ Step 6: Substitute back: $$\int x^2 e^x dx = x^2 e^x - (2x e^x - 2 e^x) + C = e^x (x^2 - 2x + 2) + C$$ (c) $\int_0^\infty \lambda x e^{-\lambda x} dx$ Step 1: Let $I = \int_0^\infty \lambda x e^{-\lambda x} dx$ Step 2: Substitute $u=\lambda x$, $du=\lambda dx$, $dx = du/\lambda$ Step 3: Then $$I = \int_0^\infty u e^{-u} du$$ Step 4: This is a Gamma integral $\Gamma(2) = 1! =1$ Step 5: So $$I = 1$$ (d) $\int_{-\infty}^\infty \frac{\lambda x}{2} e^{-i|x|} dx$ Step 1: The integrand is an odd function of $x$ because $x$ is odd, $|x|$ is even, so $\lambda x e^{-i|x|}$ is odd. Step 2: Integral over symmetric interval of an odd function is zero: $$0$$ 8. **Combinatorics problems:** Q1. Prove $\binom{n}{k} = \frac{n!}{k! (n-k)!}$ for positive integers $n \geq k$. -- Standard formula from definition of binomial coefficients. Q2. Split 10 students into three teams of sizes 2, 4, 4. Step 1: Number of ways: $$\frac{10!}{2!4!4!}$$ Q3. Number of ways to blend the plain cake with 5 other flavors. Step 1: If blending means selecting any subset of flavors (including empty), number of subsets of 5 flavors is $2^5=32$. Step 2: Including plain cake always, total is $32$ different types. Q4. Dice selection: (a) Select 4 dice with at least 3 red from 12 red and 6 white. Step 1: Case 3 red + 1 white: $$\binom{12}{3} \cdot \binom{6}{1}$$ Step 2: Case 4 red + 0 white: $$\binom{12}{4}$$ Step 3: Total: $$\binom{12}{3} \binom{6}{1} + \binom{12}{4}$$ (b) 4 dice with at least one red: Step 1: Total ways to choose 4 dice: $$\binom{18}{4}$$ Step 2: Subtract no red 4 dice selections (all white): $$\binom{6}{4}$$ Step 3: Total: $$\binom{18}{4} - \binom{6}{4}$$