Binomial Theorem Test

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Binomial Theorem Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

The number of terms in the expansion of $$( 1 + 5\sqrt 2 x)^9 + ( 1 -5\sqrt 2 x)^9$$ is

A
5

B
7

C
9

D
10

Solution

$$S=(1+5\sqrt{2}x)^{9}+(1-5\sqrt{2}x)^{9}$$
Let $$k=5\sqrt{2}$$
Then the above binomial expansion changes to
$$S=(1+k)^{9}+(1-k)^{9}$$

$$=[1+\:^{9}C_{1}k+\:^{9}C_{2}k^{2}+...\:^{9}C_{9}k^{9}]+[1-\:^{9}C_{1}k+\:^{9}C_{2}k^{2}-\:^{9}C_{3}k^{3}...-\:^{9}C_{9}k^{9}]$$

$$=2[1+\:^{9}C_{2}k^{2}+\:^{9}C_{4}k^{4}+\:^{9}C_{6}k^{6}+\:^{9}C_{8}k^{8}]$$
Hence in total 5 terms will be there.
Question 2
1 / -0

The sum of all the coefficients in the binomial expansion of $$(x^2 + x^3)^{319}$$ is

A
1

B
$$(2)^{318}$$

C
$$(2)^{319}$$

D
0

Solution

For the sum of the coefficients of the binomial theorem, we substitute $$x=1$$.
Hence, we get
$$(x^{2}+x^{3})^{319}|_{x=1}$$

$$=2^{319}$$
Thus sum of the coefficients of the above binomial expansion is $$2^{319}$$.
Question 3
1 / -0

Which term in the expansion of $$\left (\displaystyle \frac {x}{3}-\frac {2}{x^2}\right )^{10}$$ contains $$x^4$$?

A
1

B
3

C
4

D
5

Solution

The general term is as follows
$$T_{r+1}$$ $$=(-1)^{r}\:^{10}C_{r}x^{10-3r}3^{10-r}2^{r}$$
Hence, for the coefficient of $$x^{4}$$
$$10-3r=4$$
$$3r=6$$
$$r=2$$.
So, $$r+1=3$$rd term
Question 4
1 / -0

If $$C_0 , C_1 , C_2 .... C_n$$ denote the binomial coefficients in the expansion of $$(1 + x)^n$$ , then the value of $$\displaystyle \sum_{r=0}^n(r+1)C_r$$ is

A
$$n 2^n$$

B
$$(n+1)2^{n-1}$$

C
$$(n+2)2^{n-1}$$

D
$$(n+2)2^{n-2}$$

Solution

$$\sum(r+1)C_{r}$$
$$=\sum(r)C_{r}+\sum C_{r}$$
$$=\dfrac{d(1+x)^{n}}{dx}_{x=1}+2^{n}$$
$$=n(2)^{n-1}+2^{n}$$
$$=2^{n-1}(n+2)$$
Question 5
1 / -0

If $$C_0, C_1, C_2, ..............C_n$$ are binomial coefficients then $$\displaystyle \frac {1}{n!0!}+\frac {1}{(n-1)!1!}+\frac {1}{(n-2)!2!}+ ....+\frac {1}{0!n!}$$ is equal to

A
$$2^n$$

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

C
$$\displaystyle \frac {2^n}{n!}$$

D
none of these

Solution

Multiplying and dividing the entire expression by n!, we get
$$\dfrac{1}{n!}[\dfrac{n!}{0!.n!}+\dfrac{n!}{(n-1)!.1!}+\dfrac{n!}{(n-2)!.2!}+...\dfrac{n!}{0!.n!}]$$

$$=\dfrac{1}{n!}[\:^{n}C_{0}+\:^{n}C_{1}+\:^{n}C_{2}+...\:^{n}C_{n}]$$

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

$$=\dfrac{2^{n}}{n!}$$.
Question 6
1 / -0

If $$(1+x-2x^2)^8= a_0 + a_1x + a_2x^2 +....+ a_{16}x^{16}$$ then the sum $$a_1+a_3+a_5+ .... +a_{15}$$ is equal to

A
$$-2^7$$

B
$$2^7$$

C
$$2^8$$

D
none of these

Solution

Substituting x=1, we get
$$0=a_{0}+a_{1}+a_{2}+...a_{16}$$
Substituting x=-1 we get
$$2^{8}=a_{0}-a_{1}+a_{2}+...-a_{15}+a_{16}$$
Subtracting ii from i, we get
$$-2^{8}=2[a_{1}+a_{3}+a_{5}+a_{7}+...a_{15}]$$
Or
$$a_{1}+a_{3}+a_{5}+a_{7}+...a_{15}=-2^{7}$$
Question 7
1 / -0

The number of terms which are free from radical signs in the expansion of $$(y^{\frac 15} + x^{\frac 1{10}})^{55}$$ are

A
5

B
6

C
7

D
none of these

Solution

In the expansion of $$(y^{1/5} + x^{1/10})^{55}$$,the general term is
$$T_{r+1} = ^{55}C_r (y^{1/5})^{55-r} (x^{1/10})^r = ^{55}C_r.y^{11-r/5}x^{r/10}$$
This $$T_{r+1}$$ will be independent of radicals if the exponents r/5 and r/10 are integers, for $$0\leq r \leq 55$$ which is possible only when $$r = 0, 10, 20, 30, 40, 50$$.
$$\therefore$$ There are six terms viz. $$T_1,T_{11},T_{21},T_{31},T_{41},T_{51}$$, which are independent of radicals.
Question 8
1 / -0

If the sum of the coefficients in the expansion of $$(1 -3x + 10x^2)^n$$ is $$a$$ and if the sum of the coefficients in the expansion of $$(1 + x^2)^n$$ is $$b$$, then

A
$$a = 3b$$

B
$$a = b^3$$

C
$$b=a^3$$

D
none of these

Solution

Given ,
$$a =$$ sum of the coefficient in the expansion of $$(1 - 3x + 10x^2)^n$$
$$\Rightarrow a = (1 - 3 + 10)^n = (8)^n$$ (Putting $$x = 1$$)
$$\Rightarrow a = (2)^{3n}$$
Now, $$b =$$ sum of the coefficients in the expansion of $$(1 + x^2)^n$$
$$\Rightarrow b = (1 + 1)^n = 2^n$$ (Putting $$x = 1$$)
Clearly, $$a = b^3$$
Question 9
1 / -0

$$\binom{n}{0}+2\binom{n}{1}+2^2\binom{n}{2}+ .... +2^n\binom{n}{n}$$ is equal to

A
$$2^n$$

B
$$0$$

C
$$3^n$$

D
none of these

Solution

Given, $$\binom{n}{0}+2\binom{n}{1}+2^2\binom{n}{2}+ .... +2^n\binom{n}{n}$$
Simplifying the above expression we get
$$=\:^{n}C_{0}2^{0}+\:^{n}C_{1}2^{1}+\:^{n}C_{2}2^{2}+...\:^{n}C_{n}2^{n}$$
Consider $$x=2$$ in the expansion of $$(1+x)^{n}$$.
$$=(1+2)^{n}$$
$$=3^{n}$$.
Question 10
1 / -0

If $$(2x+3x^2)^6 = a_0 + a_1x + a_2x^2+.....+ a_{12} x^{12}$$, then values of $$a_0$$ and $$a_6$$ are

A
$$0, 6$$

B
$$0, 2^6$$

C
$$1, 6$$

D
$$0$$

Solution

Writing the general term we get

$$T_{r+1}$$
$$=\:^{6}C_{r}x^{6+r}2^{6-r}.3^{r}$$

Hence coefficient of $$x^{0}$$ is 0.

Thus $$a_{0}=0$$.
For $$a_{6}$$ r=0.

Hence

$$\:^{6}C_{0}.2^{6}.3^{0}$$
$$=2^{6}$$.

Hence

$$a_{6}=2^{6}$$.

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Binomial Theorem Test - 33

Result

Binomial Theorem Test - 33

Score

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Rank

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SHARING IS CARING

Question 1

5

7

9

10

Solution

Question 2

1

$$(2)^{318}$$

$$(2)^{319}$$

0

Solution

Question 3

1

3

4

5

Solution

Question 4

$$n 2^n$$

$$(n+1)2^{n-1}$$

$$(n+2)2^{n-1}$$

$$(n+2)2^{n-2}$$

Solution

Question 5

$$2^n$$

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

$$\displaystyle \frac {2^n}{n!}$$

none of these

Solution

Question 6

$$-2^7$$

$$2^7$$

$$2^8$$

none of these

Solution

Question 7

5

6

7

none of these

Solution

Question 8

$$a = 3b$$

$$a = b^3$$

$$b=a^3$$

none of these

Solution

Question 9

$$2^n$$

$$0$$

$$3^n$$

none of these

Solution

Question 10

$$0, 6$$

$$0, 2^6$$

$$1, 6$$

$$0$$

Solution

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