Sequences, Series, and the Binomial Theorem

Write the first four terms of $a_n=\dfrac{(-1)^n\, n}{n+1}$.

$a_1=\dfrac{-1}{2}$,   $a_2=\dfrac{2}{3}$,   $a_3=\dfrac{-3}{4}$,   $a_4=\dfrac{4}{5}$.

The factor $(-1)^n$ makes the signs alternate, starting negative.

Let $a_1=2$ and $a_n=3a_{n-1}-1$. Find $a_2,a_3,a_4$.

$a_2=3(2)-1=5,\quad a_3=3(5)-1=14,\quad a_4=3(14)-1=41.$

Evaluate $\displaystyle\sum_{k=1}^{5}(2k+3)$.

$\displaystyle\sum_{k=1}^{5}(2k+3)=2\sum_{k=1}^{5}k+\sum_{k=1}^{5}3 =2\cdot\frac{5\cdot 6}{2}+3\cdot 5=30+15=45.$

Check by hand: $5+7+9+11+13=45.$ ✓

For $3,7,11,15,\dots$ find $a_{20}$ and $S_{20}$.

Here $a_1=3$, $d=4$.   $a_{20}=3+(20-1)(4)=3+76=79.$

$S_{20}=\dfrac{20}{2}(a_1+a_{20})=10(3+79)=10(82)=820.$

Sum the first $6$ terms of $2,6,18,54,\dots$

Here $a_1=2$, $r=3$.   $S_6=2\cdot\dfrac{1-3^{6}}{1-3}=2\cdot\dfrac{1-729}{-2}=2\cdot\dfrac{-728}{-2}=728.$

Evaluate $\displaystyle\sum_{k=1}^{\infty} 5\left(\tfrac{2}{3}\right)^{k-1}$.

Here $a_1=5$ and $r=\tfrac{2}{3}$, and $|r|<1$, so it converges:

Prove that $\displaystyle\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$ for all integers $n\ge 1$.

Base case ($n=1$): the left side is $1$; the right side is $\dfrac{1(1+1)}{2}=1$. They agree. ✓

Inductive step: assume $\displaystyle\sum_{k=1}^{m} k=\frac{m(m+1)}{2}$ for some integer $m\ge 1$. We must show the formula holds for $m+1$:

Factor out $(m+1)$:

This is exactly the formula with $n=m+1$. By induction the statement holds for all $n\ge 1$. $\blacksquare$

Expand $(x+2)^4$.

Row of coefficients: $1,4,6,4,1$. With $y=2$:

Find the term containing $x^4$ in $(2x-1)^6$.

General term: $\displaystyle\binom{6}{k}(2x)^{6-k}(-1)^{k}$. We need power $6-k=4$, so $k=2$:

The coefficient of $x^4$ is $240$.

From $10$ students, (a) how many ways to pick a president then a vice-president? (b) how many ways to pick a $2$-person committee?

(a) Order matters: ${}_{10}P_2=\dfrac{10!}{8!}=10\cdot 9=90.$

(b) Order does not matter: ${}_{10}C_2=\dbinom{10}{2}=\dfrac{10\cdot 9}{2\cdot 1}=45.$

A bag has $5$ red and $3$ blue marbles. Two are drawn at random. Find $P(\text{both red})$.

Total ways to choose $2$ of $8$: $\dbinom{8}{2}=28.$   Favorable (both red): $\dbinom{5}{2}=10.$

Evaluate $\displaystyle\sum_{k=1}^{n}\frac{1}{k(k+1)}$ and then take $n\to\infty$.

Split each term: $\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$. Then

As $n\to\infty$, $\dfrac{1}{n+1}\to 0$, so the infinite sum is $\displaystyle\sum_{k=1}^{\infty}\frac{1}{k(k+1)}=1.$

(a) Evaluate $\displaystyle\sum_{k=1}^{\infty} k\left(\tfrac{1}{2}\right)^{k}$.

Here $r=\tfrac12$ and $|r|<1$, so

(b) Evaluate $\displaystyle\sum_{k=1}^{10} k^2$ with the closed form:

Solve $a_1=2,\ a_n=3a_{n-1}-1$ for an explicit formula, and check against the earlier hand-computed terms $2,5,14,41$.

Fixed point: $L=3L-1\Rightarrow L=\tfrac12$. So $a_n-\tfrac12=3\big(a_{n-1}-\tfrac12\big)$, a geometric sequence with ratio $3$ and first term $a_1-\tfrac12=\tfrac32$:

Check: $a_1=\tfrac{1+3}{2}=2$, $a_2=\tfrac{1+9}{2}=5$, $a_3=\tfrac{1+27}{2}=14$, $a_4=\tfrac{1+81}{2}=41.$ ✓

(a) A pizza shop offers $6$ toppings. How many topping-subsets are possible (including plain)?

Choose any subset: $\displaystyle\sum_{k=0}^{6}\binom{6}{k}=2^{6}=64.$

(b) Verify the hockey-stick identity $\displaystyle\sum_{i=2}^{5}\binom{i}{2}=\binom{6}{3}$.

Left: $\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\binom{5}{2}=1+3+6+10=20$. Right: $\binom{6}{3}=20.$ ✓

Prove $\displaystyle\sum_{k=0}^{n}\binom{n}{k}=2^{\,n}$ for all integers $n\ge 0$.

Base case ($n=0$): the left side is $\binom{0}{0}=1$ and the right side is $2^0=1$. They agree. ✓

Inductive step: assume $\displaystyle\sum_{k=0}^{m}\binom{m}{k}=2^{\,m}$. Using Pascal's rule $\binom{m+1}{k}=\binom{m}{k-1}+\binom{m}{k}$ (with $\binom{m}{-1}=\binom{m}{m+1}=0$),

This is the formula with $n=m+1$. By induction it holds for all $n\ge 0$. $\blacksquare$

Find the coefficient of $x^{2}$ in $(1+x)^{3}(1+x)^{4}$.

The product is $(1+x)^{7}$, so the coefficient of $x^2$ is $\binom{7}{2}=21$. As a Vandermonde check with $m=3,n=4,r=2$:

Find the constant term in $\left(2x+\dfrac{1}{x}\right)^{6}$.

General term: $\displaystyle\binom{6}{k}(2x)^{6-k}\!\left(\frac{1}{x}\right)^{k}=\binom{6}{k}2^{\,6-k}\,x^{\,6-2k}$. The constant term needs $6-2k=0$, so $k=3$:

The constant term is $160$.