Binomial Theorem Test

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Binomial Theorem Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

The coefficient of $$x^5$$ in the expansion of $$(1+x)^{21}+(1+x)^{22}+........+(1+x)^{30}$$ is:

A
$$^{51}C_5$$

B
$$^9C_5$$

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

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

Solution

Given :
$$(1+x)^{21}+(1+x)^{22}+(1+x)^{23}+.....+(1+x)^{30}$$

Coefficient of $$x^{5}$$

$$\ ^{21}C_{5}(1)^{16}x^{5}+\ ^{22}C_{5}(1)^{17}x^{5}+\ ^{23}C_{5}(1)^{18}x^{5}+.... +\ ^{29}C_{5}(1)^{24}x^{5}+\ ^{30}C_{5}(1)^{25}x^{5}$$

$$=x^{5}\left[\ ^{21}C_{5}+\ ^{22}C_{5}+\ ^{23}C_{5}+.....+\ ^{29}C_{5}+\ ^{30}C_{5}\right] ....... (1)$$

Now we know $$\ ^{n}C_{r}+\ ^{n}C_{r-1}=\ ^{n+1}C_{r}$$

$$\therefore$$ Adding and substracting $$\ ^{21}C_{6}$$ from eq $$(1)$$

the coefficient of $$x^{5}$$

$$\ ^{21}C_{5}+\ ^{21}C_{6}+\ ^{22}C_{5}+\ ^{25}C_{5}+.....+\ ^{29}C_{5}+\ ^{30}C_{5}-\ ^{21}C_{6}$$

$$=\ ^{22}C_{6}+\ ^{22}C_{5}+\ ^{23}C_{5}+.....+\ ^{29}C_{5}+\ ^{30}C_{5}-\ ^{21}C_{6}$$

$$=\ ^{23}C_{6}+\ ^{23}C_{5}+....+\ ^{29}C_{5}+\ ^{30}C_{5}-\ ^{21}C_{6}$$

This way
$$\ ^{30}C_{6}+\ ^{30}C_{5}-\ ^{21}C_{6}$$

$$\Rightarrow \ ^{31}C_{6}- \ ^{21}C_{6}$$ [Using $$\ ^{n}C_{r}+\ ^{n}C_{r-1}=\ ^{n+1}C_{r}$$]
Question 2
1 / -0

The co efficient of $${ x }^{ 99 }$$ in the polynomial $$\left( x-1 \right) \left( x-2 \right) \left( x-3 \right) .....\left( x-100 \right) $$ is

A
$$1$$

B
$$-5050$$

C
$$5050$$

D
$$-1$$

Solution

$${ x }^{ 99 }$$ in $$\left( x-1 \right) \left( x-2 \right) \left( x-3 \right) .....\left( x-100 \right) $$
$${ x }^{ 99 }=-$$(sum of roots)
$$=-\left( 1+2+3+...100 \right) $$
$$=-\cfrac { 100\times 101 }{ 2 } $$
$$=-5050$$
Question 3
1 / -0

If the sum of the co-efficient in the expansion of $$(a+b)^n$$ is $$1024$$, then the greatest co-efficient in the expansion is

A
$$252$$

B
$$352$$

C
$$452$$

D
$$552$$

Solution

Sum of coefficient in the expansion of $$(a+b)^{n}$$ is $$1024$$
put $$a=b=1,2^{n}=1024\implies n=10$$
The greatest coefficient is $$^{10}C_{5}=252$$
Question 4
1 / -0

If $$n\ge 2$$ then $$3.{ C }_{ 1 }-4.{ C }_{ 2 }+5.{ C }_{ 3 }-......+{ \left( -1 \right) }^{ n-1 }\left( n+2 \right) .{ C }_{ n }$$ is equal to

A
$$-1$$

B
$$2$$

C
$$-2$$

D
$$1$$

Solution

$$(1+x)^n=^nC_0+^nC_1x+^nC_2x^2+....+^nC_n x^n$$

$$x^2(1+x)^n=^nC_0 x^2+^nC_1x^3+^nC_2x^4+....+^nC_n x^{n+2}$$
differentiate w.r.t $$x$$

$$2x(1+x)^n+nx^2(1+x)^{n-1}=2^nC_0 x+3 ^nC_1x^2+4 ^nC_2x^3+....+(n+2)^nC_n x^{n+1}$$

put $$x=-1$$

$$0=-2+3 ^nC_1-4 ^nC_2+5 ^nC_3+.....+(-1)^{n-1}(n+2) ^nC_n$$

so $$3 ^nC_1- 4 ^nC_2+5 ^nC_3+.....+(-1)^{n-1}(n+2) ^nC_n=2$$
Question 5
1 / -0

For $$2\leq r \leq n, \left( ^{n+1} _r\right)+\left( ^n _{r-1}\right) + \left( ^n _{r-2}\right)$$ is equal to-

A
$$\left( ^{n+1} _{r}\right)$$

B
$$2\left( ^{n+1} _{r-1}\right)$$

C
$$2\left( ^{n+2} _r\right)$$

D
$$\left( ^{n+2} _{r}\right)$$

Solution

$$^{n+1}C_r +^nC_{r-1} +^nC_{r-2}\\=^{n+1}C_r+^{n+1}C_{r-1} \\=^{n+2}C_{r} $$
Question 6
1 / -0

The $$6^{th}$$ coefficient in the expansion of $$\left (2x^2 - \dfrac {1}{3x^2}\right)^{10}$$

A
$$-\dfrac{986}{27}$$

B
$$\dfrac{986}{27}$$

C
$$\dfrac{896}{27}$$

D
$$- \dfrac{896}{27}$$

Solution

$${ \left( 2{ x }^{ 2 }-\dfrac { 1 }{ { { 3x }^{ 2 } } } \right) }^{ 10 }$$
$$6^{th}$$ coefficient $$\rightarrow$$ $$5^{th}$$ term
$$={ ^{ 10 }{ C } }_{ 5 }{ \left( 2{ x }^{ 2 } \right) }^{ 5 }{ \left( \dfrac { -1 }{ { { 3x }^{ 2 } } } \right) }^{ 5 }$$
$$={ ^{ 10 }{ C } }_{ 5 }\times 2^5\times \dfrac{-1}{3^5}$$
$$=-\dfrac{252\times 32}{243}$$
$$=-\dfrac{896}{27}.$$
Hence, the answer is $$-\dfrac{896}{27}.$$
Question 7
1 / -0

The sum $$^{10}C_3 + ^{11}C_3 + ^{12}C_3 + .... + ^{20}C_3$$ is equal to

A
$$^{21}C_4$$

B
$$^{21}C_4 - ^{10}C_4$$

C
$$^{21}C_4 - ^{11}C_4$$

D
$$^{21}C_{17}$$

Solution

$$ ^{10}C_{3} +\ ^{11}C_{3} +\ ^{12}C_{3} + --- +\ ^{20}C_{3} $$ ....(i)
addition and subtraction $$^{10}C_{4}$$ in equation (i)
$$ ^{10}C_{4} -\ ^{10}C_{4} +\ ^{10}C_{3} +\ ^{11}C_{3} + --- +\ ^{20}C_{3}$$
[we know that - $$ ^nc_{r-1} +\ ^nc_{r} =\ ^{n+1}c_{r}] $$
$$^{10}C_{3} +\ ^{10}C_{4} +\ ^{11}C_{3} + --- +\ ^{20}C_{3} -\ ^{10}C_{4} $$
$$=\ ^{11}C_{4} +\ ^{11}C_{3} +\ ^{12}C_{3} + ---+\ ^{20}C_{3} -\ ^{10}C_{4} $$
$$=\ ^{12}C_{4} +\ ^{12}C_{3} + --- +\ ^{20}C_{3} -\ ^{10}C_{4} $$
$$ =\ ^{13}C_{4} +\ ^{14}C_{3} + --- +\ ^{20}C_{3} -\ ^{10}C_{4}$$
Similarly, Continue
So, we get $$^{21}C_{4} - \ ^{10}C_{4}$$
So, option (B) is correct
Question 8
1 / -0

Prove that $$C_0+C_1+C_2+.....C_n=2^n$$

A
$$2^n$$

B
$$2^{n-1}$$

C
$$2^{n+1}$$

D
None of the above.

Solution

$$(1+x)^n=^nC_0+^nC_1x+\cdots+^nC_nx^n$$

Let $$x=1$$

$$\Rightarrow C_0+C_1.1+C_2.1^2+\cdots+C_n.1^n=(1+1)^n=2^n$$

$$\Rightarrow C_0+C_1+C_2.1+\cdots+C_n=(1+1)^n=2^n$$
Question 9
1 / -0

$$\sum\limits_{k = 1}^{n - r} {{}^{n - k}\mathop C\nolimits_r = {}^x\mathop C\nolimits_y } $$

A
$$x = n + 1 ; y = r$$

B
$$x = n ; y = r + 1$$

C
$$x = n ; y = r$$

D
$$x = n + 1 ; y= r + 1$$

Solution
Question 10
1 / -0

Find the $$13^{th}$$ terms in the expansion of $$\left(9x-\dfrac{1}{3\sqrt{x}}\right)^{18}, x \neq 0$$.

A
$$18564$$

B
$$87328$$

C
$$17374$$

D
$$35546$$

Solution

$${\left( {9x - \frac{1}{{\sqrt {3x} }}} \right)^{18}}$$
$${T_{r + 1}} = {}^{18}{C_r}{\left( {9x} \right)^{18 - r}}{\left( { - \dfrac{1}{{3\sqrt x }}} \right)^r}$$
putting $$r = 12$$
$${T_{13}} = {}^{18}{C_{12}}{\left( {9x} \right)^{18 - 12}}{\left( { - \dfrac{1}{{3\sqrt x }}} \right)^r}$$
$$ = {}^{18}{C_6} \times {\left( {9x} \right)^6} \times \dfrac{1}{{{3^{12}} \times {x^6}}}$$
$$ = {}^{18}{C_6} \times {9^6} \times {x^6} \times \dfrac{1}{{{3^{12}} \times {x^6}}}$$
$$ = {}^{18}{C_6} \times {3^{12}} \times \dfrac{1}{{{3^{12}}}}$$

$$ = \dfrac{{18 \times 17 \times 16 \times 15 \times 14 \times 13}}{{6 \times 5 \times 4 \times 3 \times 2}}$$

$$ = 17 \times 4 \times 3 \times 7 \times 13$$
$$ = 18564$$

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Binomial Theorem Test - 43

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Binomial Theorem Test - 43

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Question 1

$$^{51}C_5$$

$$^9C_5$$

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

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

Solution

Question 2

$$1$$

$$-5050$$

$$5050$$

$$-1$$

Solution

Question 3

$$252$$

$$352$$

$$452$$

$$552$$

Solution

Question 4

$$-1$$

$$2$$

$$-2$$

$$1$$

Solution

Question 5

$$\left( ^{n+1} _{r}\right)$$

$$2\left( ^{n+1} _{r-1}\right)$$

$$2\left( ^{n+2} _r\right)$$

$$\left( ^{n+2} _{r}\right)$$

Solution

Question 6

$$-\dfrac{986}{27}$$

$$\dfrac{986}{27}$$

$$\dfrac{896}{27}$$

$$- \dfrac{896}{27}$$

Solution

Question 7

$$^{21}C_4$$

$$^{21}C_4 - ^{10}C_4$$

$$^{21}C_4 - ^{11}C_4$$

$$^{21}C_{17}$$

Solution

Question 8

$$2^n$$

$$2^{n-1}$$

$$2^{n+1}$$

None of the above.

Solution

Question 9

$$x = n + 1 ; y = r$$

$$x = n ; y = r + 1$$

$$x = n ; y = r$$

$$x = n + 1 ; y= r + 1$$

Solution

Question 10

$$18564$$

$$87328$$

$$17374$$

$$35546$$

Solution

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