Binomial Theorem (VSAQs)

Maths-2A | 6. Binomial Theorem – VSAQs:
Welcome to VSAQs in Chapter 6: Binomial Theorem. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

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VSAQ-1 : Find the 7^th term in (4/x³ + x²/2 )¹⁴

In $$(x+y)^n$$ $$T_{r+1} = ^nC_r x^{n-r}y^r$$

$$T_7 = T_6+1 = ^{14}C_6\left(\frac{4}{x^3}\right)^{14-6}\left(\frac{x^2}{2}\right)^6 = ^{14}C_6 \frac{48}{2^6}.x^{12}/x^{24} = ^{14}C_6 \frac{48}{64}.x^{12}/x^{24} = ^{14}C_6 \frac{3}{4}x^{-12}$$

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VSAQ-2 : Find the 4^th term from the end in the expansion of (2a+5b)⁸

The 4th term from the end in $$(2a+5b)^8 =$$ 4th term from the beginning in $$(5b+2a)^8$$

Formula: $$T_{r+1} = ^nC_r x^{n-r}y^r$$

$$T_4 = T_3+1 = ^8C_3 (5b)^{8-3} (2a)^3 = ^8C_3 (5b)^5 (2a)^3 = ^8C_3.5^5b^5.2^3a^3 = ^8C_3.2^3.5^5.a^3b^5$$

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VSAQ-3 : Find the middle term (s) in the expansion of (3x/7-2y)¹⁰

The binomial exponent $$n = 10$$ is even.

The middle term is $$T_{\frac{10}{2}+1} = T_{5+1} = T_6$$

$$T_6 = T_{5+1} = ^{10}C_5\left(\frac{3x}{7}\right)^{10-5}\left(-2y\right)^5 = – ^{10}C_5\left(\frac{3}{7}\right)^5.x^5.2^5.y^5 = – ^{10}C_5 \left(\frac{3}{7}\right)^5.x^5.y^5$$

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VSAQ-4 : Find the middle term (s) in the expansion (4x²+5x³)¹⁷

The binomial exponent $$n = 17$$ is odd.

The 2 middle terms are $$T_{\frac{17+1}{2}} = T_{\frac{18}{2}} = T_9$$ and the next term $$T_{10}$$

In $$(4x^2 + 5x^3)^{17}$$

$$T_9 = T_{8+1} = ^{17}C_8 (4x^2)^{17-8}\cdot(5x^3)^8 = ^{17}C_8 (4x^2)^9\cdot(5x^3)^8$$

$$= ^{17}C_8 4^9\cdot(x^2)^9\cdot5^8\cdot(x^3)^8 = ^{17}C_8\cdot4^9\cdot5^8\cdot x^{42}$$

Also $$T_{10} = T_{9+1} = ^{17}C_9(4x^2)^{17-9}\cdot(5x^3)^9 = ^{17}C_9 (4x^2)^8\cdot(5x^3)^9$$

$$= ^{17}C_9 4^8\cdot(x^2)^8\cdot5^9\cdot(x^3)^9 = ^{17}C_9\cdot4^8\cdot5^9\cdot x^{43}$$

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VSAQ-5 : Find the term independent of x in the expansion of (√x/3+3/2x²)¹⁰

The general term of $$\left(\frac{\sqrt{x}}{3} + \frac{3}{2\sqrt{x}^2}\right)^{10}$$ is $$T_{r+1} = ^{10}C_r \left(\frac{\sqrt{x}}{3}\right)^{10-r}\left(\frac{3}{2\sqrt{x}^2}\right)^r$$

$$^{10}C_r\left(\frac{1}{3}\right)^{10-r}\left(\frac{3}{2}\right)^r(x^{\frac{10-r}{2}})(\frac{1}{x^2})^r = ^{10}C_r\left(\frac{1}{3}\right)^{10-r}\left(\frac{3}{2}\right)^r.x^{\frac{10-r}{2}-2r} = ^{10}C_r\left(\frac{1}{3}\right)^{10-r}\left(\frac{3}{2}\right)^r.x^{5-\frac{5r}{2}}$$

Take $$10-\frac{r}{2} – 2r = 0$$ $$10-\frac{r}{2} = 2r$$

$$10-r = 4r$$

$$5r = 10$$

$$r = 2$$

From (1) independent term $$^{10}C_2 \left(\frac{1}{3}\right)^{10-2}\left(\frac{3}{2}\right)^2 = \frac{10 \times 9}{2 \times 1}\left(\frac{1}{3}\right)^4\left(\frac{3^2}{2^2}\right) = \frac{45}{4}$$

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VSAQ-6 : Find the coeff. of x¹¹ in (2x²+3/x³)¹³

The general term of $$(2x^2 + \frac{3}{x^3})^{13}$$ is $$T_{r+1} = ^{13}C_r(2x^2)^{13-r}\left(\frac{3}{x^3}\right)^r$$

$$^{13}C_r(2)^{13-r}\cdot3^r\cdot x^{26-2r}\cdot x^{-3r} = ^{13}C_r (2)^{13-r}(3)^r\cdot x^{26-5r}$$

To get the coefficient of $$x^{11}$$ we take $$26 – 5r = 11 \Rightarrow 5r = 15 \Rightarrow r = 3$$

From (1) the coefficient of $$x^{11} = ^{13}C_3 (2)^{10}(3)^3 = (286)(2^{10})(3^3)$$

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VSAQ-7 : Find the largest binomial coefficients in the expansions of
i. (1+x)²⁴
ii. (1+x)¹⁹

(i) n = 24 is even. So largest binomial coefficient = Coefficient of middle term

$$n C \frac{n}{2} = ^{24}C_{\frac{24}{2}} = ^{24}C_{12}$$

(ii) n = 19 is odd. So the largest binomial coefficients = Coefficients of the two middle most terms

$$n C \left(\frac{n-1}{2}\right) = ^{19}C_{\frac{19-1}{2}} = ^{19}C_{9}$$ and $$n C \left(\frac{n+1}{2}\right) = ^{19}C_{\frac{19+1}{2}} = ^{19}C_{10}$$

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VSAQ-8 : If ²²C_r   is the largest binomial coefficient in the expansion of (1+x)²². Find the value of ¹³C_r

The Binomial exponent n = 22 is even

The largest binomial coefficient is $$n C \frac{n}{2} = ^{22}C_{\frac{22}{2}} = ^{22}C_{11}$$

Now $$^{22}C_r = ^{22}C_{11} \Rightarrow r = 11$$

$$^{13}C_r = ^{13}C_{11} = ^{13}C_{2} = \frac{13 \times 12}{2 \times 1} = 78$$

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VSAQ-9 : If the coefficients of (2r+4)^th term and (3r+4)^th term in the expansion of (1+x)²¹ are equal, then find r

Coefficient of the r^th term in $$(1+x)^n$$ is $$nC_{r-1}$$

For the equation $$(1+x)^{21}$$ you’re equating the coefficients of the $$(2r+4)^{th}$$ term and the $$(3r+4)^{th}$$ term

$$21C_{2r+3} = 21C_{3r+3}$$

Formula $$nC_r = nC_s \Rightarrow r = s$$ (or) $$r + s = n$$

$$2r + 3 = 3r + 3$$

$$(2r+3)+(3r+3) = 21 \Rightarrow 5r + 6 = 21 \Rightarrow 5r = 15 \Rightarrow r = 3$$

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VSAQ-10 : Prove that C₀+2.C₁+4.C₂+8.C₃+…..+2ⁿ.C_n=3ⁿ

We have $$(1+x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n$$

Put x = 2 we get $$C_0 + C_1 \cdot 2 + C_2 \cdot 2^2 + \ldots + C_n \cdot 2^n = (1+2)^n = 3^n$$

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VSAQ-11 : Find the number of terms in the expansion of (2x+3y+z)⁷

The number of terms in the trinomial expansion of $$(x + y + z)^n = \frac{(n + 1)(n + 2)}{2}$$

The number of terms in the expansion of $$(2x + 3y + z)^7 = \frac{(7 + 1)(7 + 2)}{2} = 8 \times 9 / 2 = 36$$

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VSAQ-12 : Find the set E of x for which the binomial expansion (3-4x)^3/4 is valid

$$(3 – 4x)^{3/4} = 3^{3/4}(1 – 4x/3)^{3/4}$$ This is valid when $$|4x/3|<1$$

$$|x| < 3/4$$

$$x \in (-3/4, 3/4)$$

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VSAQ-13 : Find the set of ‘x’ for which the binomial expansion of (2+3x)^-2/3 is valid

$$(2 + 3x)^{-2/3} = 2^{-2/3}(1 + 3x/2)^{-2/3}$$

This is valid when $$|3x/2| < 1$$

$$|x| < 2/3$$

$$x \in (-2/3, 2/3)$$