Binomial Theorem Test

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Binomial Theorem Test - 31

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Question 1
1 / -0

If $$\displaystyle \left ( 1+x+x^{2} \right )^{25}=a_{0}+a_{1}x+a_{2}x^{2}+\cdot \cdot \cdot +a_{50}\cdot x^{50}$$ then $$\displaystyle a_{0}+a_{2}+a_{4}+\cdot \cdot \cdot +a_{50}$$ is

A
even

B
odd & of the form $$3n$$.

C
odd & of the form $$(3n-1)$$

D
Odd & of the form $$(3n+1)$$

Solution

$$(1+x+x^{2})^{25}=a_{0}+a_{1}x+a_{2}x^{2}+...a_{50}x^{50}$$

Substituting $$x=1$$, we get

$$3^{25}=a_{0}+a_{1}+a_{2}+...a_{50}$$ ...(i)

Substituting $$x=-1$$ we get

$$1=a_{0}-a_{1}+a_{2}-a_{3}...+a_{50}$$ ...(ii)

Adding $$i$$ and $$ii$$, we get

$$3^{25}+1=2[a_{0}+a_{2}+a_{4}+...a_{50}]$$

Hence

$$a_{0}+a_{2}+a_{4}+...a_{50}=\dfrac{3^{25}+1}{2}$$

Since the LHS is a positive integer hence the RHS has to be a positive integer.
Therefore

$$3^{25}+1$$ will be divisible by $$2$$.

Hence

$$3^{25}+1$$ will be even.

And

$$3^{2n+1}+1$$ is always even and is divisible by $$4$$.

Hence

$$\dfrac{3^{25}+1}{2}$$ will be even.
Question 2
1 / -0

If the expansion of $${ \left( 1+x \right) }^{ 50 }$$, the sum of coefficients of add powers of $$x$$ is

A
$${ 2 }^{ 50 }$$

B
$${ 2 }^{ 49 }$$

C
$$0$$

D
None of these

Solution

The sum of coefficients of odd powers of $$x$$
$$=_{ }^{ 50 }{ { C }_{ 1 }^{ } }+^{ 50 }{ { C }_{ 3 }^{ } }+...+^{ 50 }{ { C }_{ 49 }^{ } }={ 2 }^{ 50-1 }={ 2 }^{ 49 }$$.
Question 3
1 / -0

If $$\displaystyle \begin{pmatrix} p \\ q \end{pmatrix}$$ =0 for $$\displaystyle p< q$$ p,$$\displaystyle q\epsilon W$$ then $$\displaystyle \sum_{r=0}^{\infty}$$ $$\displaystyle \begin{pmatrix} n \\ 2r \end{pmatrix}$$

A
$$\displaystyle 2^{n}$$

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

C
$$\displaystyle 2^{2n-1}$$

D
$$\displaystyle \hat{2n}C_{n}$$

Solution

$$\sum \:^{n}C_{2r}$$
Is the sum of even odd term in the binomial expansion of $$(1+x)^{n}$$
Hence
$$\sum \:^{n}C_{2r}$$ will always be equal to $$2^{n-1}$$.
Question 4
1 / -0

The coefficients of three consecutive terms in the expansion of $${ (1+x) }^{ n }$$ are in the ratio $$5:10:21$$. find $$n$$

A
$$5$$

B
$$4$$

C
$$7$$

D
$$8$$
Question 5
1 / -0

The sum of the coefficients of all the even powers of x in the expansion of $$\displaystyle \left ( 2x^{2}-3x+1 \right )^{11}$$ is

A
$$\displaystyle 2 . 6^{10}$$

B
$$\displaystyle 3 . 6^{10}$$

C
$$\displaystyle 6^{11}$$

D
None of the above

Solution

Given equation is $$(2x^{2}-3x+1)^{11}$$
$$=(2x-1)^{11}(x-1)^{11}$$
$$=(3)^{11}.(2)^{11-1}$$
$$=3^{11}.2^{10}$$
$$=6^{10}.3$$
Question 6
1 / -0

Number of rational terms in the expansion of $$\displaystyle \left ( \sqrt{2}+\sqrt[4]{3} \right )^{100}$$ is

A
$$25$$

B
$$26$$

C
$$27$$

D
$$28$$

Solution

The general term for the following expression is

$$T_{r+1}=\:^{100}C_{r}2^{50-\tfrac{r}{2}}.3^{\tfrac{r}{4}}$$.

Hence we get rational terms for

$$r=0,4,8,12....100$$

$$a_{n}=a+(n-1).d$$

$$100=0+(n-1).4$$

$$25=n-1$$

$$n=26$$
Question 7
1 / -0

If the coefficients of $$\displaystyle x^{7}$$ & $$\displaystyle x^{8}$$ in the expansion of $$\displaystyle \left [ 2+\frac{x}{3} \right ]^{n}$$ are equal then the value of n is

A
$$15$$

B
$$45$$

C
$$55$$

D
$$56$$

Solution

Hence

$$T_{8}=T_{9}$$

$$\:^{n}C_{8}2^{n-8}3^{-8}=\:^{n}C_{7}2^{n-7}3^{-7}$$

$$\:^{n}C_{8}=6\:^{n}C_{7}.$$

$$\dfrac{n!}{(n-8)!.8!}=6.\dfrac{n!}{(n-7)!.7!}$$

$$n-7=6.8$$

$$n-7=48$$

$$n=55$$.
Question 8
1 / -0

The sum of the binomial coefficients of $$\displaystyle \left [2 x+\frac{1}{x} \right ]^{n}$$ is equal to 243. value of n is

A
5

B
6

C
7

D
8

Solution

Substituting $$x=1$$ for sum of coefficients, we get
$$3^{n}=243$$
$$3^{n}=3^{5}$$
Hence
$$n=5$$
Question 9
1 / -0

Find the $$\displaystyle 7^{th}$$ term of $$\displaystyle \left ( 3x^{2}-\frac{1}{3}\right)^{10}$$.

A
$$\displaystyle \frac{66}{3}x^{7}$$

B
$$\displaystyle \frac{70}{3}x^{7}$$

C
$$\displaystyle \frac{66}{3}x^{8}$$

D
$$\displaystyle \frac{70}{3}x^{8}$$

Solution

Writing the general term, we get
$$T_{r+1}=(-1)^{r}\:^{10}C_{r}3^{10-2r}.x^{20-2r}$$
For 7th term r is equal to 6.
Hence
$$T_{7}=(-1)^{6}\:^{10}C_{6}3^{10-12}.x^{20-12}$$

$$=\dfrac{210.x^{8}}{9}$$

$$=\dfrac{70x^{8}}{3}$$.
Question 10
1 / -0

Directions For Questions

If $$n$$ is positive integer and if $$\displaystyle \left ( 1+4x+4x^{2} \right )^{n}=\sum_{r=0}^{2n}a_{r}x^{r}$$ where $$\displaystyle a_{i}'s$$ are $$(i = 0, 1, 2, 3, ....., 2n)$$ real numbers

...view full instructions

The value of $$\displaystyle 2\sum_{r=0}^{n}a_{2r-1}$$ is

A
$$\displaystyle 9^{n}-1$$

B
$$\displaystyle 9^{n}+1$$

C
$$\displaystyle 9^{n}-2$$

D
$$\displaystyle 9^{n}+2$$

Solution

$$(1+4x+4x^{2})^{n}=a_{0}+a_{1}x+a_{2}x^{2}+...(a_{2n}x^{2n})$$
Substituting $$x=1$$ we get
$$9^{n}=a_{0}+a_{1}+a_{2}+...(a_{2n})$$
Substituting $$x=-1$$ we get
$$1=a_{0}-a_{1}+a_{2}-a_{3}...(a_{2n})$$
Adding both we get
$$2(a_{0}+a_{2}+a_{4}+...a_{2n})=9^{n}+1$$
Hence
$$2\sum _{k=0} ^{n}a_{2k}=9^{n}+1$$

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Binomial Theorem Test - 31

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Binomial Theorem Test - 31

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Question 1

even

odd & of the form $$3n$$.

odd & of the form $$(3n-1)$$

Odd & of the form $$(3n+1)$$

Solution

Question 2

$${ 2 }^{ 50 }$$

$${ 2 }^{ 49 }$$

$$0$$

None of these

Solution

Question 3

$$\displaystyle 2^{n}$$

$$\displaystyle 2^{n-1}$$

$$\displaystyle 2^{2n-1}$$

$$\displaystyle \hat{2n}C_{n}$$

Solution

Question 4

$$5$$

$$4$$

$$7$$

$$8$$

Question 5

$$\displaystyle 2 . 6^{10}$$

$$\displaystyle 3 . 6^{10}$$

$$\displaystyle 6^{11}$$

None of the above

Solution

Question 6

$$25$$

$$26$$

$$27$$

$$28$$

Solution

Question 7

$$15$$

$$45$$

$$55$$

$$56$$

Solution

Question 8

5

6

7

8

Solution

Question 9

$$\displaystyle \frac{66}{3}x^{7}$$

$$\displaystyle \frac{70}{3}x^{7}$$

$$\displaystyle \frac{66}{3}x^{8}$$

$$\displaystyle \frac{70}{3}x^{8}$$

Solution

Question 10

$$\displaystyle 9^{n}-1$$

$$\displaystyle 9^{n}+1$$

$$\displaystyle 9^{n}-2$$

$$\displaystyle 9^{n}+2$$

Solution

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