Question 1

If the three consecutive coefficients in the expansion of (1 + x) n are 28, 56 and 70, then the value of n is ______ .

Accepted Answer

8 — [{"id": "ffe461cb-4d65-785d-b598-1b9c4b717387", "type": "html", "value": " Solution: Let the three consecutive coefficients be n C r−1 = 28, n C r = 56 and n C r+1 = 70, so that n C r / n C r−1 = [n − r + 1] / [r] = 56 / 28 = 2 and n C r+1 / n C r = [n − r] / [r + 1] = 70 / 56 = 5 / 4 This gives n + 1 = 3r and 4n − 5 = 9r 4n − 5 / n + 1 = 3 ⇒ n = 8 "}]

Question 2

Find the independent term of x in expansion of (3x - (2/x 2 )) 15

Accepted Answer

5 — [{"id": "dc11eccf-3c28-c902-0f97-bc58729f9e75", "type": "html", "value": " Solution: The general term of (3x - (2/x 2 )) 15 is written, as T r+1 = 15 C r (3x) 15-r (-2/x 2 ) r . It is independent of x if, 15 - r - 2r = 0 => r = 5 "}]

Question 3

Total no. of term in the expansion of (x+y) 99 is

Accepted Answer

100 — [{"id": "385f1e78-188b-a4b0-d19c-d25dd45a45f8", "type": "html", "value": " Total number of term in the (A+B) n is n+1. Ans=100 "}]

Question 4

What is the general term in the binomial expansion of (1+2x) n

Accepted Answer

n C r 1 r (-2x) n-r — [{"id": "0f38e17a-b8bf-5e2d-a636-319dce0b949d", "type": "html", "value": " Solution "}]

Question 5

Expand each of the expressions in Exercises 1 to 5.

1. $(1 - 2x)^5$

Accepted Answer

Using the binomial theorem, (1 - 2x)^5 = \sum_{k=0}^5 \binom{5}{k} (1)^{5-k} (-2x)^k
= \sum_{k=0}^5 \binom{5}{k} (-2)^k x^k

Calculating each term:

k=0: \binom{5}{0} (-2)^0 x^0 = 1
k=1: \binom{5}{1} (-2)^1 x^1 = 5 	imes (-2) x = -10x
k=2: \binom{5}{2} (-2)^2 x^2 = 10 	imes 4 x^2 = 40x^2
k=3: \binom{5}{3} (-2)^3 x^3 = 10 	imes (-8) x^3 = -80x^3
k=4: \binom{5}{4} (-2)^4 x^4 = 5 	imes 16 x^4 = 80x^4
k=5: \binom{5}{5} (-2)^5 x^5 = 1 	imes (-32) x^5 = -32x^5

Therefore,
(1 - 2x)^5 = 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5 — The binomial theorem states that (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k.
Here, a = 1, b = -2x, n = 5.
We substitute and simplify each term to get the expansion.

Question 6

Expand each of the expressions in Exercises 1 to 5.

2. $\left(\frac{2}{x} - \frac{x}{2}\right)^5$

Accepted Answer

Let a = \frac{2}{x}, b = -\frac{x}{2}, n=5.

Using binomial theorem:
\left(\frac{2}{x} - \frac{x}{2}ight)^5 = \sum_{k=0}^5 \binom{5}{k} \left(\frac{2}{x}ight)^{5-k} \left(-\frac{x}{2}ight)^k

Calculate each term:

k=0: \binom{5}{0} (2/x)^5 (-x/2)^0 = 1 	imes \frac{32}{x^5} 	imes 1 = \frac{32}{x^5}

k=1: \binom{5}{1} (2/x)^4 (-x/2)^1 = 5 	imes \frac{16}{x^4} 	imes \left(-\frac{x}{2}ight) = 5 	imes 16 	imes \left(-\frac{1}{2x^3}ight) = -40/x^3

k=2: \binom{5}{2} (2/x)^3 (-x/2)^2 = 10 	imes \frac{8}{x^3} 	imes \frac{x^2}{4} = 10 	imes \frac{8}{x^3} 	imes \frac{x^2}{4} = 10 	imes 2 / x = \frac{20}{x}

k=3: \binom{5}{3} (2/x)^2 (-x/2)^3 = 10 	imes \frac{4}{x^2} 	imes \left(-\frac{x^3}{8}ight) = 10 	imes 4 	imes \left(-\frac{x}{8}ight) = -5x

k=4: \binom{5}{4} (2/x)^1 (-x/2)^4 = 5 	imes \frac{2}{x} 	imes \frac{x^4}{16} = 5 	imes \frac{2}{x} 	imes \frac{x^4}{16} = \frac{5x^3}{8}

k=5: \binom{5}{5} (2/x)^0 (-x/2)^5 = 1 	imes 1 	imes \left(-\frac{x^5}{32}ight) = -\frac{x^5}{32}

Therefore,
\left(\frac{2}{x} - \frac{x}{2}ight)^5 = \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32} — Apply the binomial theorem with a = 2/x and b = -x/2, expand each term carefully considering powers of x and signs.

Question 7

Expand each of the expressions in Exercises 1 to 5.

3. $(2x - 3)^6$

Accepted Answer

Using binomial theorem:
(2x - 3)^6 = \sum_{k=0}^6 \binom{6}{k} (2x)^{6-k} (-3)^k

Calculate terms:

k=0: \binom{6}{0} (2x)^6 (-3)^0 = 1 	imes 64 x^6 	imes 1 = 64 x^6
k=1: \binom{6}{1} (2x)^5 (-3)^1 = 6 	imes 32 x^5 	imes (-3) = -576 x^5
k=2: \binom{6}{2} (2x)^4 (-3)^2 = 15 	imes 16 x^4 	imes 9 = 2160 x^4
k=3: \binom{6}{3} (2x)^3 (-3)^3 = 20 	imes 8 x^3 	imes (-27) = -4320 x^3
k=4: \binom{6}{4} (2x)^2 (-3)^4 = 15 	imes 4 x^2 	imes 81 = 4860 x^2
k=5: \binom{6}{5} (2x)^1 (-3)^5 = 6 	imes 2 x 	imes (-243) = -2916 x
k=6: \binom{6}{6} (2x)^0 (-3)^6 = 1 	imes 1 	imes 729 = 729

Therefore,
(2x - 3)^6 = 64 x^6 - 576 x^5 + 2160 x^4 - 4320 x^3 + 4860 x^2 - 2916 x + 729 — Apply binomial theorem with a=2x, b=-3, n=6. Calculate each term using binomial coefficients and powers.

Question 8

Expand each of the expressions in Exercises 1 to 5.

4.  $\left(\frac{x}{3} +\frac{1}{x}\right)^5$

Accepted Answer

Let a = \frac{x}{3}, b = \frac{1}{x}, n=5.

Using binomial theorem:
\left(\frac{x}{3} + \frac{1}{x}ight)^5 = \sum_{k=0}^5 \binom{5}{k} \left(\frac{x}{3}ight)^{5-k} \left(\frac{1}{x}ight)^k

Calculate each term:

k=0: \binom{5}{0} \left(\frac{x}{3}ight)^5 \left(\frac{1}{x}ight)^0 = 1 	imes \frac{x^5}{3^5} 	imes 1 = \frac{x^5}{243}

k=1: \binom{5}{1} \left(\frac{x}{3}ight)^4 \left(\frac{1}{x}ight)^1 = 5 	imes \frac{x^4}{81} 	imes \frac{1}{x} = 5 	imes \frac{x^3}{81} = \frac{5x^3}{81}

k=2: \binom{5}{2} \left(\frac{x}{3}ight)^3 \left(\frac{1}{x}ight)^2 = 10 	imes \frac{x^3}{27} 	imes \frac{1}{x^2} = 10 	imes \frac{x}{27} = \frac{10x}{27}

k=3: \binom{5}{3} \left(\frac{x}{3}ight)^2 \left(\frac{1}{x}ight)^3 = 10 	imes \frac{x^2}{9} 	imes \frac{1}{x^3} = 10 	imes \frac{1}{9x} = \frac{10}{9x}

k=4: \binom{5}{4} \left(\frac{x}{3}ight)^1 \left(\frac{1}{x}ight)^4 = 5 	imes \frac{x}{3} 	imes \frac{1}{x^4} = 5 	imes \frac{1}{3x^3} = \frac{5}{3x^3}

k=5: \binom{5}{5} \left(\frac{x}{3}ight)^0 \left(\frac{1}{x}ight)^5 = 1 	imes 1 	imes \frac{1}{x^5} = \frac{1}{x^5}

Therefore,
\left(\frac{x}{3} + \frac{1}{x}ight)^5 = \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5} — Apply binomial theorem with a = x/3 and b = 1/x, carefully simplifying powers of x and denominators.

Question 9

Expand each of the expressions in Exercises 1 to 5.

5.  $\left(x + \frac{1}{x}\right)^6$

Accepted Answer

Let a = x, b = \frac{1}{x}, n=6.

Using binomial theorem:
\left(x + \frac{1}{x}\right)^6 = \sum_{k=0}^6 \binom{6}{k} x^{6-k} \left(\frac{1}{x}\right)^k = \sum_{k=0}^6 \binom{6}{k} x^{6-k-k} = \sum_{k=0}^6 \binom{6}{k} x^{6-2k}

Calculate terms:

k=0: \binom{6}{0} x^6 = x^6
k=1: \binom{6}{1} x^4 = 6 x^4
k=2: \binom{6}{2} x^2 = 15 x^2
k=3: \binom{6}{3} x^0 = 20
k=4: \binom{6}{4} x^{-2} = 15 x^{-2}
k=5: \binom{6}{5} x^{-4} = 6 x^{-4}
k=6: \binom{6}{6} x^{-6} = x^{-6}

Therefore,
\left(x + \frac{1}{x}\right)^6 = x^6 + 6 x^4 + 15 x^2 + 20 + 15 x^{-2} + 6 x^{-4} + x^{-6} — Using binomial theorem, powers of x simplify to x^{6-2k} because of the reciprocal term, resulting in symmetric powers.

Question 10

Using binomial theorem, evaluate each of the following:

6.  $(96)^{3}$

Accepted Answer

We write 96 as (100 - 4).

Using binomial theorem:
(100 - 4)^3 = \sum_{k=0}^3 \binom{3}{k} 100^{3-k} (-4)^k

Calculate terms:

k=0: \binom{3}{0} 100^3 (-4)^0 = 1 	imes 1,000,000 	imes 1 = 1,000,000
k=1: \binom{3}{1} 100^2 (-4)^1 = 3 	imes 10,000 	imes (-4) = -120,000
k=2: \binom{3}{2} 100^1 (-4)^2 = 3 	imes 100 	imes 16 = 4,800
k=3: \binom{3}{3} 100^0 (-4)^3 = 1 	imes 1 	imes (-64) = -64

Sum = 1,000,000 - 120,000 + 4,800 - 64 = 884,736 — Express 96 as (100 - 4) and expand using binomial theorem to simplify calculation of cube.

Question 11

Using binomial theorem, evaluate each of the following:

7.  $(102)^{5}$

Accepted Answer

Write 102 as (100 + 2).

Using binomial theorem:
(100 + 2)^5 = \sum_{k=0}^5 \binom{5}{k} 100^{5-k} 2^k

Calculate terms:

k=0: \binom{5}{0} 100^5 2^0 = 1 	imes 10^{10} 	imes 1 = 10^{10}

k=1: \binom{5}{1} 100^4 2^1 = 5 	imes 10^8 	imes 2 = 10^{9}

k=2: \binom{5}{2} 100^3 2^2 = 10 	imes 10^6 	imes 4 = 4 	imes 10^7

k=3: \binom{5}{3} 100^2 2^3 = 10 	imes 10^4 	imes 8 = 8 	imes 10^5

k=4: \binom{5}{4} 100^1 2^4 = 5 	imes 10^2 	imes 16 = 8,000

k=5: \binom{5}{5} 100^0 2^5 = 1 	imes 1 	imes 32 = 32

Sum = 10,000,000,000 + 1,000,000,000 + 40,000,000 + 800,000 + 8,000 + 32 = 11,040,808,032 — Express 102 as (100 + 2) and expand using binomial theorem to calculate the fifth power.

Question 12

Using binomial theorem, evaluate each of the following:

8.  $(101)^{4}$

Accepted Answer

Write 101 as (100 + 1).

Using binomial theorem:
(100 + 1)^4 = \sum_{k=0}^4 \binom{4}{k} 100^{4-k} 1^k

Calculate terms:

k=0: \binom{4}{0} 100^4 = 1 	imes 10^8 = 100,000,000
k=1: \binom{4}{1} 100^3 = 4 	imes 1,000,000 = 4,000,000
k=2: \binom{4}{2} 100^2 = 6 	imes 10,000 = 60,000
k=3: \binom{4}{3} 100^1 = 4 	imes 100 = 400
k=4: \binom{4}{4} 100^0 = 1 	imes 1 = 1

Sum = 100,000,000 + 4,000,000 + 60,000 + 400 + 1 = 104,060,401 — Express 101 as (100 + 1) and expand using binomial theorem to calculate the fourth power.

Question 13

Using binomial theorem, evaluate each of the following:

9.  $(99)^{5}$

Accepted Answer

Write 99 as (100 - 1).

Using binomial theorem:
(100 - 1)^5 = \sum_{k=0}^5 \binom{5}{k} 100^{5-k} (-1)^k

Calculate terms:

k=0: \binom{5}{0} 100^5 = 1 	imes 10^{10} = 10,000,000,000
k=1: \binom{5}{1} 100^4 (-1) = 5 	imes 10^8 	imes (-1) = -500,000,000
k=2: \binom{5}{2} 100^3 1 = 10 	imes 10^6 = 10,000,000
k=3: \binom{5}{3} 100^2 (-1) = 10 	imes 10^4 	imes (-1) = -100,000
k=4: \binom{5}{4} 100^1 1 = 5 	imes 100 = 500
k=5: \binom{5}{5} 100^0 (-1) = 1 	imes (-1) = -1

Sum = 10,000,000,000 - 500,000,000 + 10,000,000 - 100,000 + 500 - 1 = 9,510,900,499 — Express 99 as (100 - 1) and expand using binomial theorem to calculate the fifth power.

Question 14

Using binomial theorem, indicate which number is larger  $(1.1)^{10000}$  or 1000.

Accepted Answer

We compare (1.1)^{10000} and 1000.

Note that 1.1 = 1 + 0.1.

Using binomial theorem,
(1 + 0.1)^{10000} = \sum_{k=0}^{10000} \binom{10000}{k} (1)^{10000-k} (0.1)^k > \binom{10000}{1} (0.1) = 10000 	imes 0.1 = 1000

Since the sum includes the first term 1 and many positive terms, (1.1)^{10000} is much larger than 1000.

Hence, (1.1)^{10000} > 1000. — Using binomial expansion, the first order term itself is 1000, and higher order terms add more, so (1.1)^{10000} > 1000.

Question 15

Find  $(a + b)^4 - (a - b)^4$ . Hence, evaluate  $(\sqrt{3} + \sqrt{2})^4 - (\sqrt{3} - \sqrt{2})^4$ .

Accepted Answer

First, expand (a + b)^4 and (a - b)^4 using binomial theorem:

(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

(a - b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4

Subtracting,
(a + b)^4 - (a - b)^4 = (a^4 - a^4) + (4a^3b + 4a^3b) + (6a^2b^2 - 6a^2b^2) + (4ab^3 + 4ab^3) + (b^4 - b^4)
= 8a^3b + 8ab^3 = 8ab(a^2 + b^2)

Now, evaluate for a = \sqrt{3}, b = \sqrt{2}:

ab = \sqrt{3} 	imes \sqrt{2} = \sqrt{6}

a^2 + b^2 = 3 + 2 = 5

Therefore,
(\sqrt{3} + \sqrt{2})^4 - (\sqrt{3} - \sqrt{2})^4 = 8 	imes \sqrt{6} 	imes 5 = 40 \sqrt{6} — Subtract expansions to eliminate even power terms, leaving terms with odd powers doubled. Substitute values to get numerical result.

Binomial Theorem

Binomial Theorem — Study Notes

7.1 Introduction

7.2 Binomial Theorem for Positive Integral Indices

Pascal's Triangle

Practice Questions — Binomial Theorem

All 14 Chapters in Mathematics