Sequences and Series — Practice

Question 1

The sum of the first $n$ terms of a sequence is $S_n=2n^{2}-3n$. Find the $15$th term.

Accepted Answer

$a_n=\left(2n^{2}-3night)-\left(2(n-1)^{2}-3(n-1)ight)=4n-5$.

$a_{15}=60-5=55$.

The form of $S_n$ is quadratic with zero constant term, confirming an AP with common difference $4$.

Question 2

Insert four geometric means between $3$ and $96$, and verify their product.

Accepted Answer

$3r^{5}=96$ gives $r^{5}=32$, so $r=2$.

The means are $6,12,24,48$.

Their product is $82944$, and $(ab)^{n/2}=(288)^{2}=82944$.

Question 3

Sum $\displaystyle\sum_{r=1}^{n}\frac{1}{(2r-1)(2r+1)}$.

Accepted Answer

$\dfrac{1}{(2r-1)(2r+1)}=\dfrac12\left[\dfrac{1}{2r-1}-\dfrac{1}{2r+1}ight]$.

Telescoping: $\dfrac12\left[1-\dfrac{1}{2n+1}ight]=\dfrac{n}{2n+1}$.

At $n=1$ both the formula and the single term give $	frac13$.

Question 4

Sum the series $2+5+10+17+26+\cdots$ to $n$ terms.

Accepted Answer

Differences are $3,5,7,9$, an AP, so the term is quadratic: $a_r=r^{2}+1$.

$\displaystyle\sum_{r=1}^{n}\left(r^{2}+1ight)=\frac{n(n+1)(2n+1)}{6}+n$.

At $n=3$: $14+3=17$, matching $2+5+10$.

Question 5

Find $\displaystyle\sum_{r=1}^{\infty}\frac{r}{2^{r}}$.

Accepted Answer

$\displaystyle\sum_{r\ge1}rx^{r}=\frac{x}{(1-x)^{2}}$ for $|x|<1$.

At $x=	frac12$: $\dfrac{\frac12}{\frac14}=2$.

Partial sums $0.5,1.0,1.375,1.625$ approach $2$, which confirms the value.

Question 6

An infinite geometric progression has sum $3$ and the sum of the squares of its terms is $\dfrac{9}{2}$. Find the progression.

Accepted Answer

$\dfrac{a}{1-r}=3$ and $\dfrac{a^{2}}{1-r^{2}}=\dfrac92$.

Substituting $a=3(1-r)$: $\dfrac{9(1-r)^{2}}{(1-r)(1+r)}=\dfrac92$, so $2(1-r)=1+r$ and $r=	frac13$.

Then $a=2$, giving $2,	frac23,	frac29,\dots$

Question 7

Sum $\displaystyle\sum_{r=1}^{n}\frac{r+2}{r(r+1)\,2^{r}}$.

Accepted Answer

$\dfrac{r+2}{r(r+1)}=\dfrac{2}{r}-\dfrac{1}{r+1}$.

So the term is $\dfrac{1}{r\,2^{r-1}}-\dfrac{1}{(r+1)2^{r}}$, which is $f(r)-f(r+1)$ with $f(r)=\dfrac{1}{r\,2^{r-1}}$.

Telescoping: $1-\dfrac{1}{(n+1)2^{n}}$.

Question 8

If $a$, $b$, $c$ are positive and $a+b+c=1$, show that $\dfrac1a+\dfrac1b+\dfrac1c\ge9$.

Accepted Answer

By the mean inequality, $(a+b+c)\left(\dfrac1a+\dfrac1b+\dfrac1cight)\ge9$.

Since $a+b+c=1$, the second bracket is at least $9$.

Equality when $a=b=c=	frac13$, which satisfies the constraint, so $9$ is attained.

Question 9

For positive reals with $x+y+z=6$, find the greatest value of $xyz$.

Accepted Answer

$\dfrac{x+y+z}{3}\ge\sqrt[3]{xyz}$, so $2\ge\sqrt[3]{xyz}$ and $xyz\le8$.

Equality when $x=y=z=2$, which satisfies the constraint.

Greatest value $8$.

Question 10

If $x^{2}+y^{2}+z^{2}=9$, find the range of $2x-y+2z$.

Accepted Answer

$(2x-y+2z)^{2}\le\left(4+1+4ight)\left(x^{2}+y^{2}+z^{2}ight)=81$.

So $-9\le2x-y+2z\le9$.

Both endpoints are attained, at $(x,y,z)$ proportional to $\pm(2,-1,2)$ with the right scale, so the range is $[-9,9]$.

Question 11

If $a$, $b$, $c$ are in AP, $b$, $c$, $d$ are in GP and $c$, $d$, $e$ are in HP, prove that $a$, $c$, $e$ are in GP.

Accepted Answer

AP: $b=\dfrac{a+c}{2}$. GP: $c^{2}=bd$. HP: $d=\dfrac{2ce}{c+e}$.

Substituting the first and third into the second: $c^{2}=\dfrac{a+c}{2}\cdot\dfrac{2ce}{c+e}$.

Cancelling $c$: $c(c+e)=e(a+c)$, so $c^{2}=ae$.

Hence $a$, $c$, $e$ are in GP.

Question 12

Find the minimum value of $\dfrac{a^{2}}{1+b}+\dfrac{b^{2}}{1+a}$ for positive $a$ and $b$ with $a+b=2$.

Accepted Answer

$\dfrac{a^{2}}{1+b}+\dfrac{b^{2}}{1+a}\ge\dfrac{(a+b)^{2}}{2+a+b}=\dfrac{4}{4}=1$.

Equality needs $\dfrac{a}{1+b}=\dfrac{b}{1+a}$, that is $a=b=1$, which satisfies the constraint.

Minimum value $1$.

Practice — Sequences and Series

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