10 Most FAQ’s of Binomial Theorem Chapter in Inter 2nd Year Maths-2A (TS/AP)

7 Marks

LAQ-1 : If the 2^nd , 3^rd and 4^th terms in the expansion of (a+x)ⁿ are respectively 240, 720 and 1080, then find the value of a,x and n

Second term in $$(a + x)^n$$ is $$T_2 = T_1+1 = n C 1 a^{n-1} X^1 = 240$$

Third term in $$(a + x)^n$$ is $$T_3 = T_2+1 = n C 2 a^{n-2} x^2 = 720$$

Fourth term in $$(a + x)^n$$ is $$T_4 = T_3+1 = n C 3 a^{n-3} x^3 = 1080$$

$$(2)/(1) \Rightarrow n C 2 a^{n-2} x^2/n C 1 a^{n-1} x = 720/240 = 3$$

$$\Rightarrow (n C 2/n C 1)(a^{n-2})(x^2)/(a^{n-1})(x) = 3$$

$$\Rightarrow (n – 1)\frac{x}{2a} = 3$$

$$\Rightarrow (n – 1)x = 6a$$

$$(3)/(2) \Rightarrow n C 3 a^{n-3} x^3/n C 2 a^{n-2} x^2 = 1080/720 = 9/6 = 3/2$$

$$\Rightarrow (n C 3/n C 2)(a^{n-3})(x^3)/(a^{n-2})(x^2) = 3/2$$

$$\Rightarrow \frac{(n – 2)x}{3a} = \frac{3}{2}$$

$$\Rightarrow 2(n – 2)x = 9a$$

$$(5)/(4) \Rightarrow \frac{2(n – 2)x}{(n – 1)x} = \frac{9a}{6a} = \frac{3}{2}$$

$$\Rightarrow 4n – 8 = 3n – 3$$

$$\Rightarrow n = 5$$

Now from (4) $$\Rightarrow (5 – 1)x = 6a \Rightarrow 4x = 6a \Rightarrow x = \frac{3a}{2}$$

Also, from (1) $$\Rightarrow n C 1 a^{n-1} x^1 = 240 \Rightarrow 5a^4(3a/2) = 240$$

$$\Rightarrow 5a^5 = 480 \Rightarrow a^5 = 96$$

$$\Rightarrow a^4 \cdot a = 240$$

$$\Rightarrow 5a^4\left(\frac{3a}{2}\right) = 240$$

$$\Rightarrow a^5 = 32$$

$$\Rightarrow a = 2$$

From (6) $$x = \frac{3a}{2} = \frac{3 \times 2}{2} = 3$$

Therefore, $$a = 2$$ $$x = 3$$ $$n = 5$$

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LAQ-2 : If the coefficients of 4 consecutive terms in the expansion of (1+x)ⁿ are a₁,a₂,a₃,a₄ respectively, then show that a₁/(a₁+a₂)+a₃/(a₃+a₄) = 2a₂/(a₂+a₃)

We take the coefficients of 4 consecutive terms $$(1 + x)^n$$ as follows:

$$a_1 = n C r$$ $$a_2 = n C r+1$$ $$a_3 = n C r+2$$ $$a_4 = n C r+3$$

L.H.S $$\frac{a_1}{a_1} + \frac{a_2 + a_3}{a_3 + a_4} = \frac{n C r}{n C r} + \frac{n C r+1 + n C r+2}{n C r+2 + n C r+3}$$

$$\frac{n C r}{n+1 C r+1} + \frac{n C r+2}{n+1 C r+3}$$ $$n C r + n C r+1 = (n+1) C r+1$$

$$\frac{n C r}{(n + 1/r + 1) n C r} + \frac{n C r+2}{(n + 1/r + 3) n C r+2}$$ $$n C r = (n/r) n-1 C r-1$$

$$\frac{r + 1}{n + 1} + \frac{r + 3}{n + 1} = \frac{r + 1 + r + 3}{n + 1} = \frac{2r + 4}{n + 1} = \frac{2(r + 2)}{n + 1}…..(1)$$

R.H.S $$\frac{2a_2}{a_2 + a_3} = \frac{2(n C r+1)}{n C r+1 + n C r+2} = \frac{2(n C r+1)}{(n+1) C r+2} = \frac{2(n C r+1)}{(n + 1/r + 2)(n C r+1)} = \frac{2}{n+1/r+2} = \frac{2(r+2)}{n+1}…(2)$$

From (1) & (2),

L.H.S = R.H.S

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LAQ-3 : If the coefficients of r^th,(r+1)^th,(r+2)^th terms in the expansion of (1+x)ⁿ are in A.P then show that n²-(4r+1)n+4r²-2=0

The coefficient of r^th, (r+1)^th, (r+2)^th terms in (1+x)ⁿ are ⁿC_r−1, ⁿC_r, ⁿC_r+1

Given that $$n C r-1 n C r n C r+1$$ are in A.P

$$n C r = n C r-1 + n C r+1$$

$$2 = \frac{n C r-1}{n C r} + \frac{n C r+1}{n C r}$$

$$2 = \frac{r}{n – r +1} + \frac{n – r}{r + 1}$$

$$2 = \frac{r(r+1) + (n-r)(n-r+1)}{(n-r+1)(r+1)}$$

$$2(n-r+1)(r+1) = r(r+1)+(n-r)(n-r+1)$$

$$2nr+2n-2r^2-2r+2r+2 = r^2+r+n^2-nr+n-nr+r^2+r$$

$$n^2-4nr-n+4r^2-2=0$$

$$n^2-n(4r+1)+4r^2-2=0$$

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LAQ-4 : If the coefficient of x¹⁰ in the expansion of (ax²+1/bx )¹¹ is equal to the coefficient of x^-10 in the expansion of (ax-1/bx²)¹¹, find the relation between a and b where a and b are real numbers

In $$(ax^2 + \frac{1}{bx})^{11}$$ the general term is $$T_{r+1} = 11 C r (ax^2)^{11-r} (\frac{1}{bx})^r$$

$$= 11 C r a^{11-r} b^{-r} x^{22-3r}….(1)$$

Put $$22-3r = 10$$

$$\Rightarrow 3r = 12$$

$$\Rightarrow r = 4$$

From (1) the coefficient of $$x^{10} \text{ coefficient is } 11 C 4 a^{11-4} b^{-4} = 11 C 4 a^7 b^{-4}….(2)$$

In $$(ax – \frac{1}{bx^2})^{11}$$ the general term is $$T_{r+1} = 11 C r (ax)^{11-r} (- \frac{1}{bx^2})^r$$

$$= (-1)^r 11 C r a^{11-r} b^{-r} x^{11-2r}….(3)$$

Put $$11 – 2r = -10$$

$$\Rightarrow 2r = 21$$

$$\Rightarrow r = 7$$

From (3) the coefficient of $$x^{-10} \text{ is } (-1)^7 11 C 7 a^{11-7} b^{-7} = – 11 C 7 a^4 b^{-7}….(4)$$

Given that the two coefficients are equal

From (2) & (4) we have

$$11 C 4 a^7 b^{-4} = – 11 C 7 a^4 b^{-7}$$

$$\Rightarrow a^7 b^{-4} = – a^4 b^{-7}$$

$$\Rightarrow a^3 = – b^{-3}$$

$$\Rightarrow a^3 b^3 = -1$$

$$\Rightarrow ab = -1$$

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LAQ-5 : Prove that C₀.C_r+C₁.C_r+1+C₂.C_r+2+…+C_n-r.C_n=²ⁿC_(n+r) for 0≤r≤n Hence deduce that
i.C₀²+C₁²+C₂²+…+C_n²=²ⁿC_n
ii.C₀C₁+C₁C₂+C₂C₃+…+C_n-1.C_n = ²ⁿC_n+1

Method-I

We have $$(1+x)^n = C_0 + C_1x + C_2x^2 + … + C_rx^r + C_{r+1}x^{r+1} + C_{r+2}x^{r+2} + … + C_nx^n….(1)$$

$$(x+1)^n = C_0x^n + C_1x^{n-1} + C_2x^{n-2} + … + C_{n-r}x^r + …. + C_n…(2)$$

Multiplying (2) and (1) we get

$$(C_0x^n + C_1x^{n-1} + C_2x^{n-2} + … + C_{n-r}x^r + C_n)(C_0 + C_1x + C_2x^2 + … + C_rx^r + C_{r+1}x^{r+1} + C_{r+2}x^{r+2} + … + C_nx^n) = (x+1)^n(1+x)^n = (1+x)^{2n}$$

Comparing the coefficient of $$x^{n+r}$$ both sides we get

$$C_0C_r + C_1C_{r+1} + C_2C_{r+2} + …. + C_{n-r}C_n = 2n C_{n+r}$$

(i) On Substituting $$r = 0$$ we get $$C_0^2 + C_1^2 + C_2^2 + … + C_n^2 = 2n C_n$$

(ii) On Substituting $$r = 1$$ $$C_0C_1 + C_1C_2 + C_2C_3 + … + C_{n-1}C_n = 2n C_{n+1}$$

Method-II

We know that $$(1+x)^n = C_0 + C_1.x + C_2.x^2 + …. + C_n.x^n…..(1)$$

On replacing x by 1/x in (1) we get $$(1 + 1/x)^n = C_0 + C_1/x + C_2/x^2 + … + C_n/x^n…..(2)$$

On multiplying (2) and (1)

$$(1 + 1/x)^n.(1+x)^n = (C_0 + C_1/x + C_2/x^2 + … + C_{n-r}/x^{n-r} … + C_n/x^n)(C_0 + C_1x + …C_rx^r + C_{r+1}x^{r+1} + C_{r+2}x^{r+2} + … + C_nx^n)….(3)$$

The coefficient of x^r in R.H.S of (3) = $$C_0C_r + C_1C_{r+1} + C_2C_{r+2} + … + C_{n-r}C_n….(4)$$

L.H.S of (3) is $$(1 + 1/x)^n (1+x)^n = (1+x)^{2n}/x^n$$

Coefficient of x^r in $$(1+x)^{2n}/x^n =$$ the coefficient of $$x^{n+r}$$ $$(1+x)^{2n} = 2n C_{n+r}….(5)$$

Hence from (4) and (5) we get

$$C_0C_r + C_1C_{r+1} + C_2C_{r+2} + … + C_{n-r}C_n = 2n C_{n+r}$$

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LAQ-6 : If n is a positive integer and x is any nonzero real number, then prove that C₀+C₁  x/2+C₂  x²/3+C₃  x³/4+…+C_n  xⁿ/(n+1)=(1+x)ⁿ⁺¹ -1)/(n+1)x

Let $$S = C_0 + \frac{C_1 \cdot x}{2} + \frac{C_2 \cdot x^2}{3} + \ldots + \frac{C_n \cdot x^n}{n+1}$$

$$= nC_0 + \frac{nC_1 \cdot x}{2} + \frac{nC_2 \cdot x^2}{3} + \ldots + \frac{nC_n \cdot x^n}{n+1}$$

$$xS = nC_0 \cdot x + \frac{nC_1 \cdot x^2}{2} + \frac{nC_2 \cdot x^3}{3} + \ldots + \frac{nC_n \cdot x^{n+1}}{n+1}$$

$$(n+1)xS = \frac{n+1}{1} \cdot nC_0 \cdot x + \frac{n+1}{2} \cdot nC_1 \cdot x^2 + \frac{n+1}{3} \cdot nC_2 \cdot x^3 + \ldots + \frac{n+1}{n+1} \cdot nC_n \cdot x^{n+1}$$

$$= (n+1)C_1 \cdot x + (n+1)C_2 \cdot x^2 + (n+1)C_3 \cdot x^3 + \ldots + (n+1)C_{n+1} \cdot x^{n+1}$$

$$(n+1)xS = (1+x)^{n+1} – 1$$

$$S = \frac{(1+x)^{n+1} – 1}{(n+1) \cdot x}$$

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LAQ-7 : Find the sum of the infinite series 1+1/3+1.3/3.6+1.3.5/3.6.9+…

Let $$S = 1 + \frac{1}{3} + \frac{1 \cdot 3}{3 \cdot 6} + \frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9} + \ldots$$ up to $$\infty$$

$$= 1 + \frac{1}{1!} \left(\frac{1}{3}\right) + \frac{1 \cdot 3}{2!} \left(\frac{1}{3}\right)^2 + \frac{1 \cdot 3 \cdot 5}{3!} \left(\frac{1}{3}\right)^3 + \ldots$$

Comparing the above series with

$$1 + \frac{p}{1!} \left(\frac{x}{q}\right) + \frac{p(p+q)}{2!} \left(\frac{x}{q}\right)^2 + \ldots = (1-x)^{-\frac{p}{q}}$$

we get $$p = 1$$ $$p + q = 3$$

$$\Rightarrow 1 + q = 3$$

$$\Rightarrow q = 2$$

Also, we have $$\frac{x}{q} = \frac{1}{3}$$

$$\Rightarrow x = \frac{q}{3} = \frac{2}{3}$$

$$S = (1-x)^{-\frac{p}{q}} = \left(1 – \frac{2}{3}\right)^{-\frac{1}{2}} = \left(\frac{1}{3}\right)^{-\frac{1}{2}}$$

$$= \left(\frac{3}{1}\right)^{\frac{1}{2}} = 3^{\frac{1}{2}} = \sqrt{3}$$

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LAQ-8 : Find the sum of the series 1-4/5+4.7/5.10-4.7.10/5.10.15+…

Let $$S = 1 – \frac{4}{5} + \frac{4 \cdot 7}{5 \cdot 10} – \frac{4 \cdot 7 \cdot 10}{5 \cdot 10 \cdot 15} + \ldots$$

$$= 1 – \frac{4}{1} \left(\frac{1}{5}\right) + \frac{4 \cdot 7}{1 \cdot 2} \left(\frac{1}{5}\right)^2 – \frac{4 \cdot 7 \cdot 10}{1 \cdot 2 \cdot 3} \left(\frac{1}{5}\right)^3 + \ldots$$

Comparing the above series with

$$1 – \frac{p}{1!} \left(\frac{x}{q}\right) + \frac{p(p+q)}{2!} \left(\frac{x}{q}\right)^2 – \ldots = (1+x)^{-\frac{p}{q}}$$

we get $$p = 4$$ $$p + q = 7$$

$$\Rightarrow 4 + q = 7$$

$$\Rightarrow q = 3$$

Also, we have $$\frac{x}{q} = \frac{1}{5}$$

$$\Rightarrow x = \frac{q}{5} = \frac{3}{5}$$

$$S = (1+x)^{-\frac{p}{q}} = \left(1 + \frac{3}{5}\right)^{-\frac{4}{3}}$$

$$= \left(\frac{8}{5}\right)^{-\frac{4}{3}} = \left(\frac{5}{8}\right)^{\frac{4}{3}} = 5^{\frac{4}{3}} / 8^{\frac{4}{3}}$$

$$= \sqrt[3]{5^4} / \sqrt[3]{8^4} = \sqrt[3]{625} / \sqrt[3]{4096} = \sqrt[3]{625} / 16$$

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LAQ-9 : If x = 1/5+1.3/5.10+1.3.5/5.10.15+…+∞, then find 3x²+6x

Given $$x = \frac{1}{5} + \frac{1 \cdot 3}{5 \cdot 10} + \frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15} + \ldots$$ to $$\infty$$

Adding 1 on both sides, we get

$$1 + x = 1 + \frac{1}{5} + \frac{1 \cdot 3}{5 \cdot 10} + \frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15} + \ldots$$

$$= 1 + \frac{1}{1!} \left(\frac{1}{5}\right) + \frac{1 \cdot 3}{2!} \left(\frac{1}{5}\right)^2 + \frac{1 \cdot 3 \cdot 5}{3!} \left(\frac{1}{5}\right)^3 + \ldots$$

Comparing the above series with

$$1 + \frac{p}{1!} \left(\frac{y}{q}\right) + \frac{p(p+q)}{2!} \left(\frac{y}{q}\right)^2 + \ldots = (1-y)^{-\frac{p}{q}}$$

We get $$p = 1$$ $$p + q = 3$$

$$\Rightarrow q = 2$$ and $$\frac{y}{q} = \frac{1}{5}$$

$$\Rightarrow y = \frac{q}{5} = \frac{2}{5}$$

$$1 + x = (1-y)^{-\frac{p}{q}} = \left(1 – \frac{2}{5}\right)^{-\frac{1}{2}} = \left(\frac{3}{5}\right)^{-\frac{1}{2}} = \left(\frac{5}{3}\right)^{\frac{1}{2}} = \sqrt{\frac{5}{3}}$$

$$\Rightarrow (1+x)^2 = \frac{5}{3}$$

$$\Rightarrow 1 + 2x + x^2 = \frac{5}{3}$$

$$\Rightarrow 3 + 6x + 3x^2 = 5$$

$$\Rightarrow 3x^2 + 6x – 2 = 0$$

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LAQ-10 : If x = 1.3/3.6+1.3.5/3.6.9+1.3.5.7/3.6.9.12+… then prove that 9x²+24x=11

Given that $$x = \frac{1.3}{3.6} + \frac{1.3.5}{3.6.9} + \frac{1.3.5.7}{3.6.9.12} + \ldots = \frac{1.3}{2!}\left(\frac{1}{3}\right)^2 + \frac{1.3.5}{3!}\left(\frac{1}{3}\right)^3 + \frac{1.3.5.7}{4!}\left(\frac{1}{3}\right)^4 + \ldots$$

Adding $$1 + \frac{1}{3}$$ on both sides, we get

$$1 + \frac{1}{3} + x = 1 + \frac{1}{1!}\left(\frac{1}{3}\right) + \frac{1.3}{2!}\left(\frac{1}{3}\right)^2 + \frac{1.3.5}{3!}\left(\frac{1}{3}\right)^3 + \ldots$$

Comparing the above series with $$1 + \frac{p}{1!}\left(\frac{y}{q}\right) + \frac{p(p+q)}{2!}\left(\frac{y}{q}\right)^2 + \ldots = (1-y)^{-\frac{p}{q}}$$

We get, $$p = 1$$ $$p + q = 3$$

$$\Rightarrow 1 + q = 3$$

$$\Rightarrow q = 2$$

Also $$\frac{y}{q} = \frac{1}{3}$$

$$\Rightarrow y = \frac{q}{3} = \frac{2}{3}$$

$$1 + \frac{1}{3} + x = (1-y)^{-\frac{p}{q}} = \left(1 – \frac{2}{3}\right)^{-\frac{1}{2}} = \left(\frac{1}{3}\right)^{-\frac{1}{2}} = \sqrt{3}$$

$$\Rightarrow \frac{4}{3} + x = \sqrt{3}$$

$$\Rightarrow x = \sqrt{3} – \frac{4}{3}$$

$$\Rightarrow 3x = 3\sqrt{3} – 4$$

$$\Rightarrow 3x + 4 = 3\sqrt{3}$$

$$\Rightarrow (3x + 4)^2 = (3\sqrt{3})^2$$

$$\Rightarrow 9x^2 + 24x + 16 = 27$$

$$\Rightarrow 9x^2 + 24x – 11 = 0$$