Binomial Theorem Test

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Binomial Theorem Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

The co-efficient of x in the expansion of $$(1-2x^3 + 3x^5) \left( 1 + \dfrac{1}{x}\right)^8$$ is :

A
56

B
65

C
154

D
62

Solution

$$ (1-2x^{3}+3x^{5})(1+\frac{1}{x})^{8} $$
$$ (1-2x^{3}+3x^{5})(^{8}C_{0}+^{8}C_{1}\frac{1}{x}+^{8}C_{2}+^{8}C_{3}\frac{1}{x^{3}}+...^{8}C_{4}) $$
$$ = 3.^{8}C_{4}-2^{8}C_{2} $$
$$ = \dfrac{3.8!}{4.4!}-2\times \dfrac{8!}{6!2!} $$
$$ = \dfrac{3\times 8\times 7\times 6\times 5\times 4!}{4\times 3\times 2\times 4!}-\dfrac{2\times 8\times 7\times 6!}{6!\times 2!} $$
$$ = 210-56 $$
$$ = 154 $$
Question 2
1 / -0

for $$x \in R,x \ne \, - 1$$ , if $${(1 + x)^{2016}} + x{(1 + x)^{2015}} + {x^2}{(1 + x)^{2014}}$$
then $${a_{17}}$$ is equal to :

A
$$\dfrac{{2017!}}{{17!2000!}}$$

B
$$\dfrac{2007!}{17!2000!}$$

C
$$\dfrac{{2017!}}{{2000!}}$$

D
$$\dfrac{{2016!}}{{16!}}$$

Solution

$$\sum_{i=0}^{2016}{{C}_{i}{x}^{i}}={\left(1+x\right)}^{2016}+x{\left(1+x\right)}^{2015}+{x}^{2}{\left(1+x\right)}^{2014}+...+{x}^{2016}$$ is in G.P

First term$$={\left(1+x\right)}^{2016}$$

Common ratio$$=\dfrac{x{\left(1+x\right)}^{2015}}{{\left(1+x\right)}^{2016}}=\dfrac{x}{1+x}$$

$$\sum_{i=0}^{2016}{{C}_{i}{x}^{i}}=\dfrac{{\left(1+x\right)}^{2016}\left(1-{\left(\dfrac{x}{1+x}\right)}^{2017}\right)}{1-\dfrac{x}{1+x}}$$

$$=\dfrac{\dfrac{{\left(1+x\right)}^{2016}}{1}-\dfrac{{\left(1+x\right)}^{2017}}{1+x}}{\dfrac{x+1-x}{1+x}}$$

$$=\dfrac{{\left(1+x\right)}^{2017}-{x}^{2017}}{1}$$

$$\therefore\,{a}_{17}=^{2017}C_{17}=\dfrac{2007!}{17!2000!}$$
Question 3
1 / -0

The co-efficient of $${x^5}$$ in the expansion of $${\left( {1 + x} \right)^{21}} + {\left( {1 + x} \right)^{22}} + ........ + {\left( {1 + x} \right)^{30}}$$ is:

A
$$^{51}{C_5}$$

B
$$^{9}{C_5}$$

C
$$^{31}{C_6}{ - ^{21}}{C_6}$$

D
$$^{30}{C_5}{ + ^{20}}{C_5}$$

Solution

According to question,

$$(1+x)^{21}$$+$$(1+x)^{22}$$+..................+$$(1+x)^{30}$$

For , coeff. of $$x^5$$ in the expansion, we need to take out coeff. of $$x^5$$ from every terms of given series.

i.e; $$^{21}C_5+^{22}C_5+.................+^{30}C_5$$

Add & subtract $$^{21}C_6$$ in this,

$$^{21}C_6+^{21}C_5+^{22}C_5+.................+^{30}C_5-^{21}C_6$$

Since, ($$^nC_r+^nC{r-1}=^{n+1}C_r$$)

Therefore,

=> $$^{22}C_6+^{22}C_5+.................+^{30}C_5-^{21}C_6$$

=> $$^{23}C_6+^{23}C_5+.................+^{30}C_5-^{21}C_6$$

Similarly , we get

=> $$^{30}C_6+^{30}C_5-^{21}C_6$$

=> $$^{31}C_6-^{21}C_6$$
Question 4
1 / -0

If $${S_n} = \sum\limits_{r = 0}^n {\dfrac{1}{{^n{C_r}}}\,\,and\,\,{t_n} = \sum\limits_{r = 0}^n {\dfrac{r}{{^n{C_r}}},\,\,then\,\,\dfrac{{{t_n}}}{{{s_n}}} = } } $$

A
$$\dfrac{1}{2}n$$

B
$$\dfrac{{2n - 1}}{2}$$

C
$$n - 1$$

D
$$2n$$

Solution
Question 5
1 / -0

In the expansion of $$(1+ax)^n$$, $$n\in N$$, then the coefficient of x and $$x^2$$ are $$8$$ and $$24$$ respectively. Then?

A
$$a=2, n=4$$

B
$$a=4, n=2$$

C
$$a=2, n=6$$

D
None of these

Solution

Here in the expansion of $$(1+x.a)^n$$,
$$n\in N$$, the $$(r+1)^{th}$$ term will be
$$T_{r+1}=\left (\dfrac {n}{r}\right)a^{n-r}. b^r$$
here first term $$a=1$$ &
second term $$b=x.a$$, so,
$$T_{r+1}=\left (\dfrac {n}{r}\right) (1)^{n-r}. (x.a)^r=\left (\dfrac {n}{r}\right).x^r.a^r$$
But given that, coefficient of $$x$$ is $$8$$ so,
For $$x$$ power $$1$$, we have to put $$r=1$$
So, $$T_2=\left (\dfrac {n}{1}\right)a^1 .x^1$$
So, $$\left (\dfrac {n}{1}\right)a=8\to (i)$$
Now, coefficient of $$x^2$$ is $$24$$, so, $$r=2$$
$$T_3=\left (\dfrac {n}{2}\right) a^2 .x^2$$
So, $$24=\left (\dfrac {n}{2}\right) a^2\to (ii)$$
From $$(1)$$ & $$(2), a.n=8\ \Rightarrow \ a=8/n$$
$$\dfrac {n(n-1)}{2}.a^2=24$$
$$\dfrac {n(n-1)}{2}\left (\dfrac {8}{n}\right)^2=24$$
$$\therefore \ \dfrac {n^2-n}{n^2}\times 32=24$$
$$\therefore \ 1-\dfrac {1}{n}=\dfrac {24}{32}=\dfrac {3}{4}$$, So, $$\boxed {n=4}$$ from $$(i)$$
$$\boxed {a=2}$$
Question 6
1 / -0

In the expansion of $$(1+x+{x}^{2}+....+{x}^{27}){(1+x+{x}^{2}+....+{x}^{14})}^{2}$$, the coefficient of $${x}^{28}$$ is

A
195

B
224

C
378

D
405

Solution
Question 7
1 / -0

The third term from the end in the expansion of $$\left( \dfrac{3x}{5}-\dfrac{5}{2x}\right)^8$$ is

A
$$\dfrac{35451}{15x^4}$$

B
$$\dfrac{45455}{16x^4}$$

C
$$\dfrac{39372}{15x^4}$$

D
$$\dfrac{39375}{16x^4}$$

Solution

$$Third\>term\>from\>the\>end\>will\>be\>\\(8-3+2)th\>term\>from\>beginning\\T_7=^8C_6(\dfrac{3x}{5})^2(\dfrac{-5}{2x})^6\\=(\dfrac{8\times\>7}{2})\times\>(\dfrac{9}{25})\times x^2\times\>(\dfrac{15625}{64\times\>x^6})\\=(\dfrac{39375}{16x^4})$$
Question 8
1 / -0

The sum of the binomial coefficients of $${ \left[ 2x+\cfrac { 1 }{ x } \right] }^{ n }$$ is equal to $$256$$. The constant term in the expansion is-

A
$$1120$$

B
$$2110$$

C
$$1210$$

D
None of the above

Solution

We know that Sum of binomial coeficients is:
$$\Rightarrow ^nC_0+^nC_1+^nC_2+....^nC_n=2^n$$

Given that

$$\Rightarrow 2^n=256$$

$$\Rightarrow n=8$$

$$\Rightarrow T_r=\space ^nC_r(2x)^{n-r}(\dfrac{1}{x})^r$$

$$\Rightarrow ^8C_r(2x)^{8-r}x^{-r}$$

$$\Rightarrow ^8C_r(2)^{8-r}x^{8-2r}$$

For contant term , the power of $$x$$ is zero.

Hence, $$\Rightarrow 8-2r=0$$

$$\Rightarrow r=4$$

$$\Rightarrow T_4=\space ^8C_4(2)^{8-4}$$

$$\Rightarrow T_4=\space ^8C_4(2)^{4}$$

$$\Rightarrow T_4=1120$$
Question 9
1 / -0

If $$\left| x \right| < 1$$ then the coefficient of $${x^n}$$ in expansion of $${\left( {1 + x + {x^2} + {x^3}....} \right)^2}$$ is

A
$$n$$

B
$$n - 1$$

C
$$n + 2$$

D
$$n + 1$$

Solution

$$(1+x+x^{2}+..... x^{n})^{2}= ^{2}C_{0}+ ^{2}C_{1} (x+x^{2}+....x^{n}) + ^{2}C_{2} (x+ ..... x^{n})^{2}$$
Coefficient of $$x^{n}= 1+ $$ coefficient of $$x^{n}$$ in $$(x+........ x^{n})^{2}$$
$$= 1+1+$$ coefficient $$x^{n}$$ in $$(x^{2}+...... x^{n})^{2}$$
$$1+1+1+$$ coefficient of $$x^{n}$$ in $$(x^{3}+....... x^{n})^{2}$$
$$=n +$$ coefficient o f$$x^{n}$$ in $$(x)^{2}$$
$$=n$$
$$A$$ is correct.
Question 10
1 / -0

$$^{10}C_0 ^2- ^{10}C_1^2 + ^{10}C_2^2- .......- (^{10}C_9) + ( ^{10}C_{10})^2= $$

A
$$0$$

B
$$(^{10}C_5)^2$$

C
$$-^{10}C_5$$

D
$$2^{10}C_5$$

Solution

$$^{10}C_0 ^2- ^{10}C_1^2 + ^{10}C_2^2- .......^{10}C_{10} ^{10}C_{10} $$
$$(1+n)^{10}=^{10}C_0+^{10}C_1 n+^{10}C_2 n^2 ....$$
$$(1-n)^{10}=^{10}C_0+^{10}C_1 n+^{10}C_2 n^2 ....$$
$$(1+n)^{10} (1-n)^{10}=$$coeffi. of $$n^{10}$$
coeff of $$n^{10}=(1-n^2)^{10}= ^{10}C_5 (-1)^5$$
$$=-^{10}C_5 $$

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Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 44

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Binomial Theorem Test - 44

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SHARING IS CARING

Question 1

56

65

154

62

Solution

Question 2

$$\dfrac{{2017!}}{{17!2000!}}$$

$$\dfrac{2007!}{17!2000!}$$

$$\dfrac{{2017!}}{{2000!}}$$

$$\dfrac{{2016!}}{{16!}}$$

Solution

Question 3

$$^{51}{C_5}$$

$$^{9}{C_5}$$

$$^{31}{C_6}{ - ^{21}}{C_6}$$

$$^{30}{C_5}{ + ^{20}}{C_5}$$

Solution

Question 4

$$\dfrac{1}{2}n$$

$$\dfrac{{2n - 1}}{2}$$

$$n - 1$$

$$2n$$

Solution

Question 5

$$a=2, n=4$$

$$a=4, n=2$$

$$a=2, n=6$$

None of these

Solution

Question 6

195

224

378

405

Solution

Question 7

$$\dfrac{35451}{15x^4}$$

$$\dfrac{45455}{16x^4}$$

$$\dfrac{39372}{15x^4}$$

$$\dfrac{39375}{16x^4}$$

Solution

Question 8

$$1120$$

$$2110$$

$$1210$$

None of the above

Solution

Question 9

$$n$$

$$n - 1$$

$$n + 2$$

$$n + 1$$

Solution

Question 10

$$0$$

$$(^{10}C_5)^2$$

$$-^{10}C_5$$

$$2^{10}C_5$$

Solution

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