MATH101 Calculus Assessed Exercise 1 Solutions

total [10] marks

The induction questions are very important;

note the importance of correct layout

A1.1. By long division we obtain $q(X)=2X+(1/2)$ and $r(X)=23X/2$;

$$\matrix{&&&&2X&+(1/2)\cr 2X^{2}+3X-4&|&4X^{3}&+7X^{2}&+5X&-2\cr&&4X^{3}&+6X^{2}&-8X&\cr&&&X^{2}&+13X&-2\cr&&&X^{2}&+(3/2)X&-2\cr&&&&23X/2&\cr}$$

so that $q(X)=2X+(1/2)$ and $r(X)=23X/2.$

[2] marks

A1.2 Let $P_{n}$ be the statement

$$1^{3}+2^{3}+\dots+n^{3}={{1}\over{4}}n^{2}(n+1)^{2}.$$

Basis of induction:$P_{1}$ asserts that $1^{3}={{1}\over{4}}.1^{1}.2^{2}$, which is true.

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

$$\eqalign{1^{3}+2^{3}+\dots+k^{3}+(k+1)^{3}&=\bigl(1^{3}+2^{3}+\dots+k^{3}\bigr)+(k+1)^{3}\cr&={{1}\over{4}}k^{2}(k+1)^{2}+(k+1)^{3}\cr\omit\span\omit\@@LTX@noalign{{\hbox{and taking out common factors we see that this is}}}\\ &={{1}\over{4}}(k+1)^{2}(k^{2}+4k+4)\cr&={{1}\over{4}}(k+1)^{2}(k+2)^{2}.\cr}$$

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

[2] marks for calculation

Total [2] for question

A1.3 Let $P_{n}$ be the statement

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

Induction step:$P_{1}$ asserts that

$${{1}\over{2}}=2-{{3}\over{2}},$$

which is true.

[1]

Induction step: Assume that $P_{n}$ holds for some $n$, and consider $P_{n+1}$; then

$$\eqalignno{\sum_{r=1}^{n+1}{{r}\over{2^{r}}}&=\sum_{r=1}^{n}{{r}\over{2^{r}}}+{{n+1}\over{2^{n+1}}}\cr&=2-{{n+2}\over{2^{n}}}+{{n+1}\over{2^{n+1}}}\cr&=2-{{2n+4}\over{2^{n+1}}}+{{n+1}\over{2^{n+1}}}\cr&=2-{{n+3}\over{2^{n+1}}},\cr}$$

hence $P_{n+1}$ holds, and hence result by induction.

[1] marks for calculation

Total [2] for question

A1.4. The partial fraction decomposition of the $n^{th}$ term is

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

so $1=A(n+2)+Bn$ and we have the pair of equations

$$1=2A,\qquad 0=A+B$$

with solutions $A=1/2$ and $B=-1/2.$ In the series we expand the sum and regroup terms; the $n^{th}$ term is

$${{1}\over{n(n+2)}}={{1}\over{2}}\Bigl({{1}\over{n}}-{{1}\over{n+2}}\Bigr)$$

[1] for partial fractions by any method

so the $N^{th}$ partial sum is

$$\eqalign{S_{N}&={{1}\over{1\cdot 3}}+{{1}\over{2\cdot 4}}+\dots+{{1}\over{N(N+2)}}\cr&={{1}\over{2}}\Bigl[\Bigl(1-{{1}\over{3}}\Bigr)+\Bigl({{1}\over{2}}-{{1}\over{4}}\Bigr)+\Bigl({{1}\over{3}}-{{1}\over{5}}\Bigr)+\dots+\Bigl({{1}\over{N-1}}-{{1}\over{N+1}}\Bigr)+\Bigl({{1}\over{N}}-{{1}\over{N+2}}\Bigr)\Bigr],\cr\omit\span\omit\@@LTX@noalign{\hbox{where the second term in each pair cancels with the first term in the pair that}}\\ \omit\span\omit\@@LTX@noalign{\hbox{is two steps to the right, so}}\\ S_{N}&={{1}\over{2}}\Bigl[1+{{1}\over{2}}-{{1}\over{N+1}}-{{1}\over{N+2}}\Bigr].\cr}$$

[2] for correct justification of this

As $N\rightarrow\infty$,

$${{1}\over{2}}\Bigl[1+{{1}\over{2}}-{{1}\over{N+1}}-{{1}\over{N+2}}\Bigr]\rightarrow{{3}\over{4}},$$

hence the sum to infinity of this series is

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

[1] for taking limit correctly

Total for question [4]

Alternatively,

$$\eqalignno{S_{N}&={{1}\over{2}}\Bigl[\Bigl(1-{{1}\over{3}}\Bigr)\cr&\quad+\Bigl({{1}\over{2}}-{{1}\over{4}}\Bigr)\cr&\quad+\Bigl({{1}\over{3}}-{{1}\over{5}}\Bigr)\cr&\quad\vdots\cr&\quad+\Bigl({{1}\over{N-1}}-{{1}\over{N+1}}\Bigr)\cr&\quad+\Bigl({{1}\over{N}}-{{1}\over{N+2}}\Bigr)\Bigr],\cr}$$

and one can cancel by knights’ moves, to get

$$S_{N}={{1}\over{2}}\Bigl[1+{{1}\over{2}}-{{1}\over{N+1}}-{{1}\over{N+2}}\Bigr].$$

The cancellation pattern suggests Knights’ moves in chess.