(Each question 2 marks)

Question 12 Marks

Find an approximation of $(0.99)^5$ using the first three terms of its expansion.

Answer

$\text { We have }(0.99)^5=(1-0.01)^5$
$={ }^5 C_0-{ }^5 C_1(0.01)+{ }^5 C_2(0.01)^2 \ldots$
$=1-0.05+0.001 \ldots$
$=0.951$

View full question & answer→

Question 22 Marks

Write the general term in the expansion of $(x^2 - yx)^{12}$, x $\ne$ 0

Answer

Here general term in the expansion of $(x^2 - yx)^{12}$ is
${T_{r + 1}}{ = ^{12}}{C_r}{({x^2})^{12 - r}}{( - yx)^r}$
$ = {( - 1)^r}^{12}{C_r}{x^{24 - 2r}} \cdot {y^r}{x^r}$$ = {( - 1)^r}^{12}{C_r}{x^{24 - r}} \cdot {y^r}$

View full question & answer→

Question 32 Marks

Write the general term in the expansion of $(x^2 - y)^6$

Answer

Here general term in the expansion of $(x^2 - y)^6$ is
${T_{r + 1}}{ = ^6}{C_r}{({x^2})^{6 - r}}{( - y)^r}$
$ = {( - 1)^r}^6{C_r}{x^{12 - 2r}}{y^r}$

View full question & answer→

Question 42 Marks

Find the coefficients of : $a^5 b^7$ in $(a - 2b)^{12}$

Answer

Here general term of the expansion $(a - 2b)^{12}$ is
$T_{r+1}=^{12}C_ra^{12-r}{(-2b)^r}$. . . (i)
Now $12 - r = 5$ and $r = 7$
$ \Rightarrow r = 12 - 5 = 7$
Putting $r = 7$ in (i)
$\therefore {T_8}{ = ^{12}}{C_7}{(a)^5}{( - 2b)^7}$${ = ^{12}}{C_7}{( - 2)^7}{a^5}{b^7}$
$\therefore $ Coefficient of $a^5 b^7$ in the expansion ${(a - 2b)^{12}}{ = ^{12}}{C_7}{( - 2)^7}$
$ = 792 \times - 128 = - 101376$

View full question & answer→

Question 52 Marks

Find the coefficients of : $x^5$ in $(x + 3)^8$

Answer

Here general term of the expansion $(x + 3)^8$ is
${T_{r + 1}}{ = ^8}{C_r}{(x)^{8 - r}}{(3)^r}$....(1)
Now 8 - r = 5 $ \Rightarrow $ r = 8 - 5 = 3 . . .
$\therefore $ Putting r = 3 in (1)
${T_4}{ = ^8}{C_3}{x^5}{(3)^3}$
Coefficient of $x^5$ on the expansion ${(x + 3)^8}{ = ^8}{C_3} \cdot {(3)^3} = 1512$

View full question & answer→

Question 62 Marks

Using binomial theorem, evaluate: $(99)^5$

Answer

$(99)^5=(100-1)^5$
Using binomial theorem, we have
$(100-1)={ }^5 C_0(100)^5+{ }^5 C_1(100)^4(-1)+{ }^5 C_2(100)^3(-1)^2+{ }^5 C_3(100)^2(-1)^3$
$+{ }^5 C_4(100)(1)^4+{ }^5 C_5(-1)^5$
$=(100)^5+5(100)^4(-1)+10(100)^3+10(100)^2(-1)^3+5(100)+(-1)^5$
$=10000000000-500000000+10000000-100000+500-1$
$=10010000500-500100001$
$=9509900499$

View full question & answer→

Question 72 Marks

Using binomial theorem, evaluate: $(101)^4$

Answer

$(101)^4 = (100 + 1)^4$
Using binomial theorem, we have
${(100 + 1)^4}{ = ^4}{C_0}{(100)^4}{ + ^4}{C_1}{(100)^3}(1)$${ + ^4}{C_2}{(100)^2}{(1)^2}{ + ^4}{C_3}(100){(1)^3}{ + ^4}{C_4}{(1)^4}$
$=(100)^4+4(100)^3+6(100)^2+4(100)+1$
$=100000000+4000000+60000+400+1$
$=104060401$

View full question & answer→

Question 82 Marks

Using binomial theorem, evaluate: $(102)^5$

Answer

$(102)^5=(100+2)^5$
Using binomial theorem, we have
$(100+2)^5={ }^5 C_0(100)^5+{ }^5 C_1(100)^4(2)+{ }^5 C_2(100)^3(2)^2$
$+{ }^5 C_3(100)^2(2)^3+{ }^5 C_4(100)(2)^4+{ }^5 C_5(2)^5$
$=(100)^5+5(100)^4(2)+10(100)^3(2)^2+10(100)^2(2)^3+5(100)(2)^4+(2)^5$
$=10000000000+1000000000+40000000+800000+8000+32$
$=11040808032$

View full question & answer→

Question 92 Marks

Using binomial theorem, evaluate: $(96)^3$

Answer

$(96)^3=(100-4)^3$
Using binomial theorem, we have
$(100-4)^3={ }^3 C_0(100)^3+{ }^3 C_1(100)^2(-4)+{ }^3 C_2(100)(-4)^2+{ }^3 C_3(-4)^3$
$=(100)^3+3.10000(-4)+3.100 .16+(-64)$
$=1000000-120000+4800-64$
$=1004800-120064=884736$

View full question & answer→

Question 102 Marks

Prove that ${\sum\limits_{r=0}^n3^r\;^nC_r=4^n}$

Answer

$\sum_{r=0}^n\;^nC_ra^{n-r}b^r=\;\left(a\;+b\right)^n\;.........\left(1\right)$
$Now,\sum_{r=0}^n3^r\;^nC_r\;=\overset n{\underset{r=0}{\sum\nolimits}}\;^nC_r\;(1)^{n-r}.3^r\;=\;(1+3)^n\;\;\;\;(by\;(1)$
$\;\;\;=\;4^n$

View full question & answer→

Question 112 Marks

Using Binomial Theorem, indicate which number is larger (1.1) ${ }^{10000}$ or 1000.

Answer

$(1.1)^{10000}=(1+0.1)^{10000}$
$=1+{ }^{10000} \mathrm{C}_1(0.1)+{ }^{10000} \mathrm{C}_2(0.1)^2+{ }^{10000} \mathrm{C}_3(0.1)^3+\ldots .$
$=1+10000(0.1)+\text { other positive numbers }$
$=1+1000+\text { other positive numbers }$
Which is greater than 1000 .
Thus (1.1)10000 > 1000

View full question & answer→

MATHS (Each question 2 marks)

Take a timed test