Binomial Theorem Test

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Binomial Theorem Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

The first three terms in the expansion of $$(1+a)^n$$ are $$t_1=1,t_2=-18,t_3=144$$.
Use the general term to determine a and n.

A
$$a=-3,n=9$$

B
$$a=-2,n=9$$

C
$$a=-3,n=8$$

D
$$a=-2,n=8$$

Solution

As we can see from option that n is positive so we proceed with the expansion
$$T_1$$ $$=$$ $$^{n}{C}_{0}$$ $$=1$$
$$T_2$$ $$=$$ $$^{n}{C}_{1}a$$ $$ =n.a=-18 $$ ....(1)
$$T_3$$ $$=$$ $$^{n}{C}_{2}a^2$$ $$=$$ $$\dfrac{(n)(n-1)}{2}$$.$$a^2$$ $$=144$$ ....(2)
$$\Rightarrow$$$${(na)(na-a)}$$.$$=288 $$
Putting the value of $$na$$ from (1), we get
$$(-18)(-18-a)$$ $$=288$$
$$\Rightarrow$$$$-18-a$$ $$= -16$$
$$\Rightarrow$$ $$-a=2$$
So, $$a= -2$$
Substitute this value of $$a$$ in equation (1), we get
$$na=-18$$
$$\Rightarrow n(-2)=-18$$
$$\Rightarrow n=9$$
Thus value of $$a$$ and $$n$$ are $$-2$$ and $$9$$.
Question 2
1 / -0

The value of $$a_0 ^2-a_1 ^2+a_2 ^2....a_{2n} ^2$$ is

A
$$a_n$$

B
$$2a_n$$

C
$$3a_n$$

D
$$\dfrac {n+1}2 (a_0+a_1+a_2...a_{2n})$$

Solution

$$(1+x)^{2n}=1+_{ }^{ 2n }{ { C }_{ 1 }x+_{ }^{ 2n }{ { C }_{ 2 }{ x }^{ 2 }+....+_{ }^{ 2n }{ { C }_{ 2n }{ x }^{ 2n } } } }$$
or,
$$(1+x)^{2n}=a_0+a_1x+a_2x^2+...+a_{2n}x^{2n}$$ ...$$(i)$$
Similarly,
$$(1-x)^{2n}=a_0-a_1x+a_2x^2+...+a_{2n}x^{2n}$$ ....$$(ii)$$
As, $$a_0=a_{2n}$$, $$a_1=a_{2n-1},...$$and so on. So, (i) and (ii) can be written as:
$$(1-x)^{2n}=a_{2n}-a_{2n-1}x+a_{2n-2}x^2+...+a_{0}x^{2n}$$
$$(1+x)^{2n}=a_0+a_1x+a_2x^2+...+a_{2n}x^{2n}$$

So, required answer is coefficient of $$x^{2n}$$ in $$(1+x)^{2n}.(1-x)^{2n}=$$ coefficient of $$x^{2n}$$ in $$(1-x^2)^{2n}$$
$${T}_{r+1}=_{ }^{2n}{C}_r(-x^2)^r$$
$$\quad\quad=_{ }^{2n}{{C}_{r}}(-1)^r(x)^{2r}$$
So, we need, $$r=n$$. Hence, the answer is $$(-1)^n.a_n$$
For $$n-even$$, the answer is option $$A$$, i.e; $$a_n$$
Question 3
1 / -0

The middle term of expansion of $$\left (\dfrac {10}{x} + \dfrac {x}{10}\right )^{10}$$

A
$$^{7}C_{5}$$

B
$$^{8}C_{5}$$

C
$$^{9}C_{5}$$

D
$$^{10}C_{5}$$

Solution

Given

Fact: middle in the expansion of $$(x+y)^n$$ will be $$\left(\dfrac{n}{2}+1\right)$$th term if $$n$$ is even

Hence $$n=10=$$ even

So middle term will be $$6$$th term

Which is given by, $$^{10}C_5\left(\dfrac{10}{x}\right)^5\left(\dfrac{x}{10}\right)^5=^{10}C_5$$
Question 4
1 / -0

The value of $$C_1 ^2+C_2 ^2....+C_n ^2$$ (where $$C_i$$ is the $$i^{th}$$ coefficient of $$(1+x)^n$$ expansion), is:

A
$$\dfrac {n^n}{n!}$$

B
$$\dfrac {2n!}{n!n!}$$

C
$$\dfrac {2n!}{n!}$$

D
$$\dfrac {n!\times2^n}{2n!}$$

Solution

$${ \left( 1+x \right) }^{ n }=^{ n }{ { C }_{ 0 } }+x\left( ^{ n }{ { C }_{ 1 } } \right) { +{ x }^{ 2 }\left( ^{ n }{ { C }_{ 2 } } \right) + }{ x }^{ 3 }\left( ^{ n }{ { C }_{ 3 } } \right) +...+{ x }^{ n }\left( ^{ n }{ { C }_{ n } } \right) \\ $$
$${ \left( x+1 \right) }^{ n }={ x }^{ n }\left( ^{ n }{ { C }_{ 0 } } \right) +{ x }^{ n-1 }\left( ^{ n }{ { C }_{ 1 } } \right) +...+{ x }^{ 0 }\left( ^{ n }{ { C }_{ n } } \right) $$
Now, $$\quad \sum _{ r=0 }^{ n }{ { \left( ^{ n }{ { C }_{ r } } \right) }^{ 2 } } $$= co-efficient of $$x^n$$ in $$(1+x)^{2n}=^{ 2n }{ { C }_{ n } }$$
Thus, $$^{ 2n }{ { C }_{ n } }={ \left( ^{ n }{ { C }_{ 0 } } \right) }^{ 2 }+{ \left( ^{ n }{ { C }_{ 1 } } \right) }^{ 2 }+...{ \left( ^{ n }{ { C }_{ n } } \right) }^{ 2 }$$, where $${ \left( ^{ n }{ { C }_{ i } } \right) }$$ is the $$i^{th}$$ coefficient of $$(1+x)^n$$ expansion.
So, $$C_1^2+C_2^2+...+C_n^2={ \left( ^{ 2n }{ { C }_{ n } } \right) }=\dfrac { 2n! }{ n!n! } $$
Question 5
1 / -0

If $$f(n)=\sum_{s=1}^n \sum_{r=s}^n \:^nC_r \:^rC_s$$, then $$f(3)=$$

A
$$27$$

B
$$19$$

C
$$1$$

D
$$5$$

Solution

Given : $$f\left( n \right) =\overset { n }{ \underset { s=1 }{ \Sigma } }\; \overset { n }{ \underset { r=s }{ \Sigma } } { ^{ n }{ C } }_{ r }{ ^{ r }{ C } }_{ s }$$
To Find : $$f\left( 3 \right) =?$$
Sol. : $$f\left( n \right) =\left\{ { ^{ n }{ C } }_{ 1 }{ ^{ 1 }{ C } }_{ 1 }+{ ^{ n }{ C } }_{ 2 }{ ^{ 2 }{ C } }_{ 1 }+{ ^{ n }{ C } }_{ 3 }{ ^{ 3 }{ C } }_{ 1 }......{ ^{ n }{ C } }_{ n }.{ ^{ n }{ C } }_{ 1 } \right\} +\left\{ { ^{ n }{ C } }_{ 2 }{ ^{ 2 }{ C } }_{ 2 }+{ ^{ n }{ C } }_{ 3 }{ ^{ 3 }{ C } }_{ 2 }+.....{ ^{ n }{ C } }_{ n }{ ^{ n }{ C } }_{ 2 } \right\} +.....$$
                       $$+\left\{ { ^{ n }{ C } }_{ n-1 }{ ^{ n-1 }{ C } }_{ n-1 }+{ ^{ n }{ C } }_{ n }{ ^{ n }{ C } }_{ n-1 } \right\} +\left\{ { ^{ n }{ C } }_{ n }{ ^{ n }{ C } }_{ n } \right\} $$
$$\Rightarrow$$ Take $$n=3$$
$$\Rightarrow f\left( 3 \right) =\left\{ { ^{ 3 }{ C } }_{ 1 }{ ^{ 1 }{ C } }_{ 1 }+{ ^{ 3 }{ C } }_{ 2 }{ ^{ 2 }{ C } }_{ 1 }+{ ^{ 3 }{ C } }_{ 3 }{ ^{ 3 }{ C } }_{ 1 } \right\} +\left\{ { ^{ 3 }{ C } }_{ 2 }{ ^{ 2 }{ C } }_{ 2 }+{ ^{ 3 }{ C } }_{ 3 }{ ^{ 3 }{ C } }_{ 2 } \right\} +\left\{ { ^{ 3 }{ C } }_{ 3 }{ ^{ 3 }{ C } }_{ 3 } \right\} $$
          $$=\left\{ 3+3.2+3 \right\} +\left\{ 3+3 \right\} +\left\{ 1 \right\} $$
          $$=19$$
Hence, the answer is $$19.$$
Question 6
1 / -0

The value of $$(n+2)C_02^{n+1}-(n+1)C_12^n+nC_22^{n-1}+....$$ is equal to:
$$(C_r=\:^nC_r)$$

A
$$4n$$

B
$$4$$

C
$$2n+4$$

D
$$4+4n$$

Solution

We know, $${ \left( 1-x \right) }^{ n }=_{ }^{ n }{ { C }_{ 0 }{ x }^{ n } }-_{ }^{ n }{ { C }_{ 1 }{ x }^{ n-1 } }+_{ }^{ n }{ { C }_{ 2 }{ x }^{ n-2 } }-...._{ }^{ n }{ { C }_{ n } }$$
Multiplying by $$x^2$$ on both sides,
$${ x }^{ 2 }{ \left( 1-x \right) }^{ n }=_{ }^{ n }{ { C }_{ 0 }{ x }^{ n+2 } }-_{ }^{ n }{ { C }_{ 1 }{ x }^{ n+1 } }+_{ }^{ n }{ { C }_{ 2 }{ x }^{ n } }-....\\ $$
Taking the derivative,
$$2x{ \left( 1-x \right) }^{ n }+n{ x }^{ 2 }{ \left( 1-x \right) }^{ n-1 }=\left( n+2 \right) { { C }_{ 0 }{ x }^{ n+1 } }-\left( n+1 \right) { { C }_{ 1 }{ x }^{ n } }+n{ { C }_{ 2 }{ x }^{ n-1 } }-....$$
Putting $$x=2$$ in LHS, we get our required result,
$$\quad4(-1)^n+n(4)(-1)^{n-1}$$
$$=4-4n$$, $$n$$ is even
$$=4n-4$$, $$n$$ is odd.

There is $$no$$ $$option$$ matching the answer.
Question 7
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If $$P_n$$ denotes the product of all the coefficients in the expansion of $$(1+x)^n$$, then $$\dfrac {P_{n+1}}{P_n}$$ is equal to:

A
$$\dfrac {(n+2)^n}{n!}$$

B
$$\dfrac {(n+1)^{n+1}}{n+1!}$$

C
$$\dfrac {(n+1)^{n+1}}{n!}$$

D
$$\dfrac {(n+1)^n}{n+1!}$$

Solution

$$(1+x)^n$$
$${ P }_{ n+1 }=_{ }^{ n+1 }{ { C }_{ 0 } }\times _{ }^{ n+1 }{ { C }_{ 1 } }\times _{ }^{ n+1 }{ { C }_{ 2 } }\times ........\times _{ }^{ n+1 }{ { C }_{ n+1 } }\\ { P }_{ n }=_{ }^{ n }{ { C }_{ 0 } }\times _{ }^{ n }{ { C }_{ 1 } }\times _{ }^{ n }{ { C }_{ 2 } }\times ........\times _{ }^{ n }{ { C }_{ n } }$$
$$\quad \prod _{ r=0 }^{ n }{ \dfrac { _{ }^{ n+1 }{ { C }_{ r } } }{ _{ }^{ n }{ { C }_{ r } } } \quad \quad \quad \quad \quad \quad \quad \left( \because _{ }^{ n+1 }{ { C }_{ n+1 }=1 } \right) } \\ =\prod _{ r=0 }^{ n }{ \dfrac { (n+1)! }{ r!(n+1-r)! } } \times \dfrac { r!(n-r)! }{ n! } \\ =\prod _{ r=0 }^{ n }{ \dfrac { (n+1) }{ \left( n-r+1 \right) } } \\ ={ \left( n+1 \right) }^{ n }\prod _{ r=0 }^{ n }{ \dfrac { 1 }{ n-r+1 } } \\ =\dfrac { { \left( n+1 \right) }^{ n } }{ \left( n+1 \right) ! } $$
Hence, the correct option is $$D$$.
Question 8
1 / -0

If $$^{n-1}C_{r}=(k^2-3)\:^nC_{r+1}$$, then $$k\:\:\epsilon$$

A
$$(-\infty,-2]$$

B
$$[2,\infty)$$

C
$$[-\sqrt 3,\sqrt 3]$$

D
$$(\sqrt 3,2]$$

Solution

Given, $$^{n-1}{C}_{r}=(k^2-3)^{n}{C}_{r+1}$$
$$\Rightarrow$$$$\dfrac{^{n-1}{C}_{r}}{^{n}{C}_{r+1}}=(k^2-3)$$ ....Using $$^{n}{C}_{x}=\dfrac{n}{x}.^{n-1}{C}_{x-1}$$
$$\Rightarrow$$$$\dfrac{r+1}{n}=(k^2-3)$$ ....As $$n>r+1$$
So, $$0<\dfrac{r+1}{n}\le1$$
Therefore, $$0<k^2-3\le1$$
$$\Rightarrow 3<k^2\le4$$
$$\Rightarrow \sqrt{3}<k\le2$$
So, k$$\in$$$$(\sqrt{3},2]$$
Question 9
1 / -0

The value of $$\displaystyle \:^{50}C_4+\sum_{r=1}^6 \:^{56-r}C_3$$ is

A
$$\:^{55}C_4$$

B
$$\:^{55}C_3$$

C
$$\:^{56}C_3$$

D
$$\:^{56}C_4$$

Solution

$$^{50}{C}_{4}$$+ $$\sum _{ r=1 }^{6 }{^{56-r}{C}_{3} }$$
$$^{50}{C}_{4}$$+$$^{50}{C}_{3}$$+$$^{51}{C}_{3}$$+.................+$$^{55}{C}_{3}$$ Using $$^{n}{C}_{r}$$+$$^{n}{C}_{r-1}$$=$$^{n+1}{C}{r}$$
$$^{51}{C}_{4}$$+$$^{51}{C}_{3}$$+.................+$$^{55}{C}_{3}$$
Similarly the series would become like this
$$^{55}{C}_{4}$$+$$^{55}{C}_{3}$$ Using $$^{n}{C}_{r}$$+$$^{n}{C}_{r-1}$$=$$^{n+1}{C}{r}$$
$$^{56}{C}_{4}$$
Question 10
1 / -0

The coefficient of $$x^{53}$$ in the following expansions.
$$\displaystyle \sum_{m=0}^{100} \,^{100}C_m(x-3)^{100-m}\cdot 2^m$$ is

A
$$^{100}C _{47}$$

B
$$^{100}C _{53}$$

C
$$^{-100}C _{53}$$

D
$$^{-100}C _{100}$$

Solution

$$\displaystyle \sum_{m=0}^{100} \, ^{100}C_m(x-3)^{100-m}\cdot 2^m$$

Above expansion can be rewritten as

$$[(x-3)+2]^{100}=(x-1)^{100}=(1-x)^{100}$$

$$\therefore x^{53}$$ will occur in $$T_{54}$$ i.e. $$54^{th}$$ term.

So, $$m=53$$

$$\Rightarrow T_{54}={ }^{100}C_{53}(-x)^{53}$$

$$\therefore $$ Required coefficient is $${ }^{-100} C_{53}$$.

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Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 38

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Binomial Theorem Test - 38

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Question 1

$$a=-3,n=9$$

$$a=-2,n=9$$

$$a=-3,n=8$$

$$a=-2,n=8$$

Solution

Question 2

$$a_n$$

$$2a_n$$

$$3a_n$$

$$\dfrac {n+1}2 (a_0+a_1+a_2...a_{2n})$$

Solution

Question 3

$$^{7}C_{5}$$

$$^{8}C_{5}$$

$$^{9}C_{5}$$

$$^{10}C_{5}$$

Solution

Question 4

$$\dfrac {n^n}{n!}$$

$$\dfrac {2n!}{n!n!}$$

$$\dfrac {2n!}{n!}$$

$$\dfrac {n!\times2^n}{2n!}$$

Solution

Question 5

$$27$$

$$19$$

$$1$$

$$5$$

Solution

Question 6

$$4n$$

$$4$$

$$2n+4$$

$$4+4n$$

Solution

Question 7

$$\dfrac {(n+2)^n}{n!}$$

$$\dfrac {(n+1)^{n+1}}{n+1!}$$

$$\dfrac {(n+1)^{n+1}}{n!}$$

$$\dfrac {(n+1)^n}{n+1!}$$

Solution

Question 8

$$(-\infty,-2]$$

$$[2,\infty)$$

$$[-\sqrt 3,\sqrt 3]$$

$$(\sqrt 3,2]$$

Solution

Question 9

$$\:^{55}C_4$$

$$\:^{55}C_3$$

$$\:^{56}C_3$$

$$\:^{56}C_4$$

Solution

Question 10

$$^{100}C _{47}$$

$$^{100}C _{53}$$

$$^{-100}C _{53}$$

$$^{-100}C _{100}$$

Solution

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