CBSE Class 11 Mathematics Binomial Theorem Worksheet Set 05

CBSE Class 11 Mathematics Worksheet - Binomial Theorem (4). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. Expand using binomial theorem expansion \( \left(\frac{2}{x} - \frac{x}{2}\right)^5 \)?
Answer: Given expansion: \( \left(\frac{2}{x} - \frac{x}{2}\right)^5 \)
\( = {}^5C_0 \left(\frac{2}{x}\right)^5 \left(\frac{x}{2}\right)^0 - {}^5C_1 \left(\frac{2}{x}\right)^4 \left(\frac{x}{2}\right)^1 + {}^5C_2 \left(\frac{2}{x}\right)^3 \left(\frac{x}{2}\right)^2 - {}^5C_3 \left(\frac{2}{x}\right)^2 \left(\frac{x}{2}\right)^3 + {}^5C_4 \left(\frac{2}{x}\right)^1 \left(\frac{x}{2}\right)^4 - {}^5C_5 \left(\frac{2}{x}\right)^0 \left(\frac{x}{2}\right)^5 \)
\( = \left(\frac{32}{x^5}\right)(1) - 5 \left(\frac{16}{x^4}\right) \left(\frac{x}{2}\right) + 10 \left(\frac{8}{x^3}\right) \left(\frac{x^2}{4}\right) - 10 \left(\frac{4}{x^2}\right) \left(\frac{x^3}{8}\right) + 5 \left(\frac{2}{x}\right) \left(\frac{x^4}{16}\right) - (1)(1)\left(\frac{x^5}{32}\right) \)
\( = \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32} \) ans.

Question. Evaluate \( (99)^5 \) using binomial theorem?
Answer: We have, \( (99)^5 = (100 - 1)^5 \)
\( = {}^5C_0 (100)^5 (1)^0 - {}^5C_1 (100)^4 (1)^1 + {}^5C_2 (100)^3 (1)^2 - {}^5C_3 (100)^2 (1)^3 + {}^5C_4 (100)^1 (1)^4 - {}^5C_5 (100)^0 (1)^5 \)
\( = 10000000000 - 5(100000000) + 10(1000000) - 10(10000) + 5(100) - 1 \)
\( = (10000000000 + 10000000 + 500) - (500000000 + 100000 + 1) \)
\( = 10010000500 - 500100001 \)
\( = 9509900499 \) ans.

Question. Find \( (a + b)^4 - (a - b)^4 \), hence evaluate \( (\sqrt{3} + \sqrt{2})^4 - (\sqrt{3} - \sqrt{2})^4 \)?
Answer: \( (a + b)^4 - (a - b)^4 \)
\( = \left[ {}^4C_0 a^4b^0 + {}^4C_1 a^3b^1 + {}^4C_2 a^2b^2 + {}^4C_3 a^1b^3 + {}^4C_4 a^0b^4 \right] - \left[ {}^4C_0 a^4b^0 - {}^4C_1 a^3b^1 + {}^4C_2 a^2b^2 - {}^4C_3 a^1b^3 + {}^4C_4 a^0b^4 \right] \)
\( = {}^4C_1 a^3b^1 + {}^4C_3 a^1b^3 + {}^4C_1 a^3b^1 + {}^4C_3 a^1b^3 \)
\( = 2({}^4C_3 a^1b^3) + 2({}^4C_1 a^3b^1) \)
\( = 8a^3b^1 + 8a^1b^3 \dots\dots\dots\dots \{ {}^4C_3 = {}^4C_1 = 4 \} \)
\( = 8ab(a^2 + b^2) \)
\( \therefore (a + b)^4 - (a - b)^4 = 8ab(a^2 + b^2) \)
Now, for \( (\sqrt{3} + \sqrt{2})^4 - (\sqrt{3} - \sqrt{2})^4 \), put \( a = \sqrt{3} \) & \( b = \sqrt{2} \)
\( = 8\sqrt{3}\sqrt{2}(3 + 2) \)
\( = 8\sqrt{6}(5) = 40\sqrt{6} \) ans.

Question. Which is larger \( (1.1)^{10000} \) or \( 1000 \)?
Answer: \( (1.1)^{10000} = (1 + 0.1)^{10000} \)
\( = {}^{10000}C_0 + {}^{10000}C_1(0.1)^1 + \dots\dots\dots\dots \)
\( = (1)(1) + 10000(0.1) + \dots\dots\dots\dots \)
\( = 1 + 1000 + \dots\dots\dots\dots (+) \text{ positive terms} \)
\( = 1001 + \dots\dots\dots\dots > 1000 \)
\( \therefore (1.1)^{10000} \) is larger than \( 1000 \) ans.

Question. Prove that \( \sum_{r=0}^n 3^r \cdot {}^nC_r = 4^n \)?
Answer: L.H.S. \( \sum_{r=0}^n 3^r \cdot {}^nC_r \)
\( = 3^0 \cdot {}^nC_0 + 3^1 \cdot {}^nC_1 + 3^2 \cdot {}^nC_2 + \dots\dots\dots\dots 3^n \cdot {}^nC_n \) ……………….. (1)
R.H.S. \( 4^n = (1 + 3)^n \)
\( = {}^nC_0 \cdot 3^0 + {}^nC_1 \cdot 3^1 + {}^nC_2 \cdot 3^2 + \dots\dots\dots\dots {}^nC_n \cdot 3^n \) ………………… (2)
From (1) and (2), L.H.S = R.H.S. (proved)

Question. Prove that \( 6^n - 5n \) always leaves the remainder 1, when divided by 25?
Answer: \( 6^n - 5n = (1 + 5)^n - 5n \)
\( = \left( {}^nC_0 \cdot 5^0 + {}^nC_1 \cdot 5^1 + {}^nC_2 \cdot 5^2 + \dots\dots\dots\dots + {}^nC_n \cdot 5^n \right) - 5n \)
\( = \left( 1 + 5n + {}^nC_2 \cdot 5^2 + {}^nC_3 \cdot 5^3 + \dots\dots\dots\dots + 5^n \right) - 5n \)
\( = 1 + {}^nC_2 \cdot 5^2 + {}^nC_3 \cdot 5^3 \dots\dots\dots\dots 5^n \)
\( = 6^n - 5n = 1 + 25 \left( {}^nC_2 + 5 \cdot {}^nC_3 + \dots\dots\dots\dots + 5^{n-2} \right) \)
Clearly it leaves the remainder 1 when divided by 25.

Question. Show that \( 9^{n+1} - 8n - 9 \) is divisible by 64?
Answer: \( 9^{n+1} - 8n - 9 \)
\( = (1 + 8)^{n+1} - 8n - 9 \)
\( = {}^{n+1}C_0 \cdot 8^0 + {}^{n+1}C_1 \cdot 8^1 + {}^{n+1}C_2 \cdot 8^2 + {}^{n+1}C_3 \cdot 8^3 + \dots\dots\dots\dots 8^{n+1} - 8n - 9 \)
\( = \left( 8n + 9 + {}^{n+1}C_2 \cdot 8^2 + {}^{n+1}C_3 \cdot 8^3 + \dots\dots\dots\dots 8^{n+1} \right) - 8n - 9 \)
\( = {}^{n+1}C_2 \cdot 8^2 + {}^{n+1}C_3 \cdot 8^3 + \dots\dots\dots\dots 8^{n+1} \)
\( = 8^2 \left( {}^{n+1}C_2 \cdot 8 \cdot n + {}^{n+1}C_3 + \dots\dots\dots\dots 8^{n+1} \right) \)
Clearly it is divisible by 64 ans.

Question. Prove that \( (a - b) \) is a factor of \( a^n - b^n \)?
Answer: \( a^n - b^n = \left[ b + (a - b) \right]^n - b^n \)
\( = {}^nC_0 \cdot b^n(a - b)^0 + {}^nC_1 \cdot b^{n-1}(a - b)^1 + {}^nC_2 \cdot b^{n-2}(a - b)^2 + \dots\dots\dots\dots {}^nC_n \cdot b^0(a - b)^n - b^n \)
\( = b^n + n \cdot b^{n-1}(a - b) + {}^nC_2 \cdot b^{n-2}(a - b)^2 + \dots\dots\dots\dots (a - b)^n - b^n \)
\( = n \cdot b^{n-1}(a - b) + {}^nC_2 \cdot b^{n-2}(a - b)^2 + \dots\dots\dots\dots (a - b)^n \)
\( = (a - b)\left[ n \cdot b^{n-1} + {}^nC_2 \cdot b^{n-2}(a - b)^2 + \dots\dots\dots\dots (a - b)^{n-1} \right] \)
Clearly it is divisible by \( (a - b) \) ans.

Question. If 'O' be the sum of odd terms & 'E' be the sum of even terms in the expansion of \( (x + a)^n \) prove that, \( O^2 - E^2 = (x^2 - a^2)^n \).
Answer: \( (x + a)^n = {}^nC_0 \cdot x^na^0 + {}^nC_1 \cdot x^{n-1}a^1 + {}^nC_2 \cdot x^{n-2}a^2 + \dots\dots\dots\dots + {}^nC_n \cdot x^0a^n \)
\( \Rightarrow (x + a)^n = \left( {}^nC_0 \cdot x^na^0 + {}^nC_2 \cdot x^{n-2} + {}^nC_4 \cdot x^{n-4}a^4 + \dots\dots\dots\dots \right) + \left( {}^nC_1 \cdot x^{n-1}a^1 + {}^nC_3 \cdot x^{n-3}a^3 + {}^nC_5 \cdot x^{n-5}a^5 + \dots\dots\dots\dots \right) \)
\( \Rightarrow (x + a)^n = O + E \) ……………… (1)
Now, \( (x - a)^n = {}^nC_0 \cdot x^na^0 - {}^nC_1 \cdot x^{n-1}a^1 + {}^nC_2 \cdot x^{n-2}a^2 - \dots\dots\dots\dots {}^nC_n \cdot x^0a^n \)
\( = \left( {}^nC_0 \cdot x^na^0 + {}^nC_2 \cdot x^{n-2}a^2 + {}^nC_4 \cdot x^{n-4}a^4 + \dots\dots\dots\dots {}^nC_n \right) - \left( {}^nC_1 \cdot x^{n-1}a^1 + {}^nC_3 \cdot x^{n-3}a^3 + {}^nC_5 \cdot x^{n-5}a^5 + \dots\dots\dots\dots \right) \)
\( \Rightarrow (x - a)^n = O - E \) ……………… (2)
Multiply eq. (1) & eq. (2)
\( (x + a)^n(x - a)^n = (O + E)(O - E) \)
\( \Rightarrow ((x + a)(x - a))^n = O^2 - E^2 \)
\( \Rightarrow (x^2 - a^2)^n = O^2 - E^2 \) (proved)

Question. Find the coefficient of \( a^4 \) in the product \( (1 + 2a)^4(2 - a)^5 \) using binomial theorem?
Answer: Since, expansion are in product
\( \therefore \) general term cannot be formed
\( (1 + 2a)^4(2 - a)^5 \)
\( = \left[ {}^4C_0 (2a)^0 + {}^4C_1 (2a)^1 + {}^4C_2 (2a)^2 + {}^4C_3 (2a)^3 + {}^4C_4 (2a)^4 \right] \times \left[ {}^5C_0 (2)^5(a)^0 - {}^5C_1 (2)^4(a)^1 + {}^5C_2 (2)^3(a)^2 - {}^5C_3 (2)^2(a)^3 + {}^5C_4 (2)^1(a)^4 - {}^5C_5 (2)^0(a)^5 \right] \)
\( = \left( 1 + 8a + 24a^2 + 32a^3 + 160a^4 \right) \times \left( 32 - 80a + 80a^2 - 40a^3 + 10a^4 - a^5 \right) \)
The terms containing \( a^4 \) are
\( = 1(10a^4) + (8a)(-40a^3) + (24a^2)(80a^2) + (32a^3)(-80a) + (16a^4)(32) \)
\( = 10a^4 - 320a^4 + 1920a^4 - 2560a^4 + 512a^4 = -438a^4 \)
\( \therefore \) coefficient of \( a^4 = -438 \) ans.

Question. Expand using binomial theorem \( \left(1 + \frac{x}{2} - \frac{2}{x}\right)^4 \)?
Answer: \( \left(1 + \frac{x}{2} - \frac{2}{x}\right)^4 = \left[ 1 + \left(\frac{x}{2} - \frac{2}{x}\right) \right]^4 \)
\( = {}^4C_0 \left(\frac{x}{2} - \frac{2}{x}\right)^0 + {}^4C_1 \left(\frac{x}{2} - \frac{2}{x}\right)^1 + {}^4C_2 \left(\frac{x}{2} - \frac{2}{x}\right)^2 + {}^4C_3 \left(\frac{x}{2} - \frac{2}{x}\right)^3 + {}^4C_4 \left(\frac{x}{2} - \frac{2}{x}\right)^4 \)
\( = (1)(1) + 4 \left(\frac{x}{2} - \frac{2}{x}\right) + 6 \left(\frac{x^2}{4} - \frac{4}{x^2} - 2\right) + 4 \left[ \frac{x^3}{8} - \frac{8}{x^3} - 3 \left(\frac{x^2}{4}\right)\left(\frac{2}{x}\right) \right] + 3 \left(\frac{x}{2}\right)\left(\frac{4}{x^2}\right) \left[ {}^4C_0 \left(\frac{x}{2}\right)^4\left(\frac{2}{x}\right)^0 - {}^4C_1 \left(\frac{x}{2}\right)^3\left(\frac{2}{x}\right)^1 + {}^4C_2 \left(\frac{x}{2}\right)^2\left(\frac{2}{x}\right)^2 - {}^4C_3 \left(\frac{x}{2}\right)^1\left(\frac{2}{x}\right)^3 + {}^4C_4 \left(\frac{x}{2}\right)^0\left(\frac{2}{x}\right)^4 \right] \dots\dots\dots\dots \left\{ \text{in } \left(\frac{x}{2} - \frac{2}{x}\right)^4 \text{ we use again binomial theorem} \right\} \)
\( = 1 + 2x - \frac{8}{x} + \frac{3x^2}{2} + \frac{24}{x^2} - 12 + \frac{x^3}{2} - \frac{32}{x^3} - 6x + \frac{24}{x} + \frac{x^4}{16} - x^2 + 6 - \frac{16}{x^2} + \frac{16}{x^4} \)
\( = \frac{16}{x} + \frac{8}{x^2} - \frac{32}{x^3} + \frac{16}{x^4} - 4x + \frac{x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} - 5 \) ans.

Question. If 'O' be the sum of odd terms & 'E' be the sum of even terms in the expansion of \( (x + a)^n \) prove that,
1. \( 4OE = (x + a)^{2n} - (x - a)^{2n} \)
2. \( 2(O^2 + E^2) = (x + a)^{2n} + (x - a)^{2n} \)
Answer:
1. We know that, \( 4OE = (O + E)^2 - (O - E)^2 \)
\( \Rightarrow 4OE = \left[(x + a)^n\right]^2 - \left[(x - a)^n\right]^2 \) …………………. {using eq. (1) and eq. (2)}
\( \Rightarrow 4OE = (x + a)^{2n} - (x - a)^{2n} \) (proved)

Squaring eq. (1) & eq. (2)
2. We get, \( (x + a)^{2n} = (O + E)^2 \) ………………. (3)
And \( (x - a)^{2n} = (O - E)^2 \) ………………. (4)
Now, eq. (3) + (4)
\( \Rightarrow (x + a)^{2n} + (x - a)^{2n} = 2(O^2 + E^2) \) (proved)