MATH101 Calculus Workshop Exercise Solutions 1

W1.1. The long division algorithm is as follows.

$$\matrix{&&&2X^{2}&+11X&+27\cr X-3&|&2X^{3}&+5X^{2}&-6X&-4\cr&&2X^{3}&-6X^{2}&&\cr&&&11X^{2}&-6X&-4\cr&&&11X^{2}&-33X&\cr&&&&27X&-4\cr&&&&27X&-81\cr&&&&&77\cr}$$

Hence the remainder is $77$.

(ii) The remainder theorem gives

$$f(X)=(X-3)q(X)+f(3)$$

where the remainder is

$$\eqalignno{f(3)&=2\cdot 3^{3}+5\cdot 3^{2}-6\cdot 3-4\cr&=54+45-18-4\cr&=77,\cr}$$

confirming the previous result.

W1.2. We have

$${{X^{3}-5X+6}\over{X^{2}+3X+2}}=X-3+{{2X+12}\over{X^{2}+3X+2}}.$$

The computation can be written in the following array.

$$\eqalign{&\quad X-3\cr X^{2}+3X+2\quad|&\quad X^{3}\qquad\quad-5X+6\cr&\quad X^{3}+3X^{2}+2X\cr&\qquad-3X^{2}-7X+6\cr&\qquad-3X^{2}-9X-6\cr&\qquad\qquad 2X+12\cr}$$

W1.3. (i) Let $P_{n}$ be the statement

$$3+7+\dots+(4n-1)=n(2n+1).$$

Basis of induction:$P_{1}$ asserts that $3=1(2+1)$, which is true.

Induction step: suppose that $P_{k}$ holds for some $k\geq 1$, and consider $P_{k+1}$. We have, by $P_{k}$,

$$\eqalign{3+7+\dots+(4k-1)+(4k+3)&=\bigl(3+7+\dots+(4k-1)\bigr)+(4k+3)\cr&=k(2k+1)+4k+3\cr&=2k^{2}+5k+3\cr&=(k+1)(2k+3).\cr}$$

hence $P_{k+1}$ holds; hence $P_{n}$ holds for all $n\geq 1$ by induction.

(ii) Let $P_{n}$ be the statement

$$1\cdot 2+2\cdot 3+\dots+n(n+1)={{1}\over{3}}n(n+1)(n+2).$$

Basis of induction:$P_{1}$ asserts that $1.2={{1}\over{3}}.1.2.3$, which is true.

Induction step: suppose that $P_{k}$ holds for some integer $k\geq 1$ and consider $P_{k+1}.$ We have, by $P_{k}$,

$$\eqalign{1\cdot 2+2\cdot 3+\dots k(k+1)+(k+1)(k+2)&=\bigl(1\cdot 2+2\cdot 3+\dots k(k+1)\bigr)+(k+1)(k+2)\cr&={{1}\over{3}}k(k+1)(k+2)+(k+1)(k+2)\cr&={{1}\over{3}}(k+1)(k+2)(k+3).\cr}$$

Hence $P_{k+1}$ holds; hence $P_{n}$ holds for all $n\geq 1$ by induction.

W1.4. Let $P_{n}$ be the statement

$$a+ar+ar^{2}+\dots ar^{n-1}=a{{1-r^{n}}\over{1-r}}.$$

Basis of induction.$P_{1}$ asserts that

$$a=a{{1-r}\over{1-r}},$$

which is true.

Induction step. Suppose that $P_{k}$ holds for some $k$ and consider $P_{k+1}$. We have, by $P_{k}$,

$$\eqalign{a+ar+\dots ar^{k-1}+ar^{k}&=\bigl(a+ar+\dots+ar^{k-1}\bigr)+ar^{k}\cr&=a{{1-r^{k}}\over{1-r}}+ar^{k}\cr&=a{{1-r^{k}+r^{k}-r^{k+1}}\over{1-r}}\cr&=a{{1-r^{k+1}}\over{1-r}}.\cr}$$

Hence $P_{k+1}$ holds; hence $P_{n}$ holds for all natural integers $n$ by induction.

W1.5. The roots of $f(X)=0$ are $-1,2,-3$ and $4$. If we happen to spot $-1$ as a root, then we proceed to factorize $f(X)$ step-by-step as

$$\eqalignno{f(X)&=X^{4}-2X^{3}-13X^{2}+14X+24\cr&=(X+1)(X^{3}-3X^{2}-10X+24)\cr&=(X+1)(X-2)(X^{2}-X-12)\cr&=(X+1)(X-2)(X+3)(X-4).\cr}$$

W1.6. Let $P_{n}$ be the statement

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

Basis of induction:$P_{1}$ asserts that $1=(1/2)(-1)^{0}2$, which is true.

Induction step: suppose that $P_{n}$ holds for some $n\geq 1$, and consider $P_{n+1}$. We have, by $P_{n}$,

$$\eqalign{\sum_{r=1}^{n+1}(-1)^{r-1}r^{2}&=\sum_{r=1}^{n}(-1)^{r-1}r^{2}+(-1)^{n}(n+1)^{2}\cr&={{(-1)^{n-1}}\over{2}}n(n+1)+(-1)^{n}(n+1)^{2},\cr&={{(-1)^{n}}\over{2}}(n+1)\bigl(-n+2(n+1)\bigr)\cr&={{(-1)^{n}}\over{2}}(n+1)(n+2)\cr}$$

hence $P_{n+1}$ holds; hence $P_{n}$ holds for all $n\geq 1$ by induction.

W1.7

$$\eqalignno{1&\qquad 1\qquad 1\qquad 1\qquad\quad 1\,\,\,\,\,\qquad 1\cr 1&\qquad 2\qquad 3\qquad 4\quad\qquad 5\qquad\,\,\,\,6\cr 1&\qquad 3\qquad 6\qquad 10\qquad\,\,15\,\,\,\qquad 21\cr 1&\qquad 4\qquad 10\qquad 20\qquad 35\qquad 56\cr 1&\qquad 5\qquad 15\qquad 35\qquad 70\qquad 126\cr 1&\qquad 6\qquad 21\qquad 56\qquad 126\qquad 252\cr}$$

W1.8(Sums of squares).

We recall that $\sum_{k=1}^{n}k^{2}=6^{-1}n(n+1)(2n+1)$, so

$$\eqalignno{\sum_{k=1}^{n}(3k^{2}-5)&=3\sum_{k=1}^{n}k^{2}-5\sum_{k=1}^{n}1\cr&={{1}\over{2}}n(n+1)(2n+1)-5n\cr&={{1}\over{2}}n\Bigl((n+1)(2n+1)-10\Bigr)\cr&={{1}\over{2}}n(2n^{2}+3n-9)\cr&={{1}\over{2}}n(n+3)(2n-3).\cr}$$

W1.9.(Geometric series)

This is a geometric series with $a=x^{2}$ and $r=x^{2}$, so $0\leq r<1$ and

$${{a}\over{1-r}}={{x^{2}}\over{1-x^{2}}}.$$

W1.10 The $n^{th}$ term of the series can be written as partial fractions

$${{1}\over{n(n+1)(n+2)}}={{A}\over{n}}+{{B}\over{n+1}}+{{C}\over{n+2}}$$

where

$$1=A(n+1)(n+2)+Bn(n+2)+Cn(n+1)$$

so we have coefficients

$$\eqalignno{n^{2}:&0=A+B+C\cr n:&0=3A+2B+C\cr 1:&1=2A\cr}$$

hence

$$\eqalignno{0&=A+B+C\cr 0&=2A+B\cr 1&=2A\cr}$$

so

$$A=1/2,\quad B=-1,\quad C=1/2;$$

$${{1}\over{n(n+1)(n+2)}}={{1/2}\over{n}}+{{-1}\over{n+1}}+{{1/2}\over{n+2}}.$$

We can write the sum to $n$ terms as

$$\eqalignno{{{1}\over{1\cdot 2\cdot 3}}&=\quad{{1/2}\over{1}}+{{-1}\over{2}}+{{1/2}\over{3}}\cr+{{1}\over{2\cdot 3\cdot 4}}&\qquad+{{1/2}\over{2}}+{{-1}\over{3}}+{{1/2}\over{4}}\cr+{{1}\over{3\cdot 4\cdot 5}}&\qquad+{{1/2}\over{3}}+{{-1}\over{4}}+{{1/2}\over{5}}\cr+{{1}\over{4\cdot 5\cdot 6}}&\qquad+{{1/2}\over{4}}+{{-1}\over{5}}+{{1/2}\over{6}}\cr\vdots&\qquad\vdots\cr+{{1}\over{(n-1)\cdot n\cdot(n+1)}}&\qquad+{{1/2}\over{n-1}}+{{-1}\over{n}}+{{1/2}\over{n+1}}\cr+{{1}\over{n\cdot(n+1)\cdot(n+2)}}&\qquad+{{1/2}\over{n}}+{{-1}\over{n+1}}+{{1/2}\over{n+2}}\cr}$$

so that when we add up, the terms on the cross diagonals with three terms cancel, leaving only the terms in the top left corner and the bottom right corner; so the $n^{th}$ partial sum is

$${{1}\over{1\cdot 2\cdot 3}}+{{1}\over{2\cdot 3\cdot 4}}+{{1}\over{3\cdot 4\cdot 5}}+\dots+{{1}\over{n\cdot(n+1)\cdot(n+2)}}$$

$$={{1}\over{2}}-{{1}\over{4}}+{{-1/2}\over{n+1}}+{{1/2}\over{n+2}};$$

and as $n\rightarrow\infty$, we obtain the sum

$${{1}\over{1\cdot 2\cdot 3}}+{{1}\over{2\cdot 3\cdot 4}}+{{1}\over{3\cdot 4\cdot 5}}+\dots={{1}\over{4}}.$$