Binomial Theorem Important Questions Mathematics

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Very Short Answer Questions [1 Mark Questions]

Ques. What is the general term of (x² - y)⁶?

Ans. General term for (x² - y)⁶ can be given by:

T_r+1=6_Cr(x²)^6−r(−y)^r

=(−1)^r6_Crx^12−2ry^r

Ques. Expand (m+n)a where r is a non-negative integer.

Ans. To expand: (m+n)^a = ^aC₀m^a + ^aC₁m^a-1n + ^aC₂m^a-2n² + ^aC₃m^a-3n³ +........+ ^aC_rm^a-ry^r +..... + ^aC₀x⁰y^a

Ques. In the expansion (a+b)ⁿ, highlight the total number of coefficients present?

Ans. It is quite clear that the total number of terms is equivalent to 1+ power of expression. As is already claimed, the power is n, which simply means that the total number of coefficients will be n+1. 

Ques. Find the 4th term in (3x - y)⁷.

Ans. Since we are aware,

Now, to determine the 4th term, we can consider r = 3, a = 3x, b = -y and n = 7

Therefore,

T₄ = T³+1 = 7C³(3x)7-3.(-y)³

⇒ -2835x4y³

Hence, it can be said that the 4^th term in (3x - y)⁷ is -2835x4y³.

Ques. Expand the following: (x/3 + 2/y)⁴

Ans. For (x/3 + 2/y)⁴:

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Short Answer Questions [2 Marks Questions]

Ques. Expand the expression (1-2x)⁵ using binomial theorem.

Ans. We know that 

Considering the following elements, x = 1, y = -2x and n = 5

(1-2x)5 = ⁵C₀1⁵ + ⁵C₁1⁴(-2x)¹ + ⁵C₂1³(-2x)² + ⁵C₃1²(-2x)³ + ⁵C₄1¹(-2x)⁴ + ⁵C₀1⁰(-2x)5

⇒ 1 - 5(2x) + 10(4x²) - 10(8x³) + 5(16x⁴) - (32x⁵)

⇒ 1 - 10x + 40x - 80x³ + 80x⁴ - 32x⁵ 

Therefore, the expanded form of (1-2x)5 is 1 - 10x + 40x - 80x³ + 80x⁴ - 32x⁵

Ques. Find the term or coefficient of x(x+1/x)10 which is independent of x.

Ans. We know that, T_r+1 = ⁿC_rx^n-r.y^r

Hence, Tr + 1 = ¹⁰C_r(x)10^-r.(1/x)r

⇒ ¹⁰Cr(x)10^-r.(x)^-r

⇒ ¹⁰Cr(x)10-2r

Put 10 - 2r = 0 we get, 

r = 5

Therefore, the term independent of x is 10C⁵.

Ques. (√2 + 1)⁵ + (√2 − 1)⁵

Ans. Here, we can see

(x + y)⁵ + (x – y)⁵ = 2[⁵C₀  x⁵ + ⁵C₂  x³ y²  +  ⁵C₄ xy⁴]

= 2(x⁵ + 10 x³ y² + 5xy⁴)

Thus, (√2 + 1)⁵ + (√2 − 1)⁵ = 2[(√2)⁵ + 10(√2)³(1)² + 5(√2)(1)⁴]

= 58√2

Ques. Determine the number of terms which are free from the radical sign in the expansion of (√5 + ⁴√n)¹⁰⁰.

Ans. T_r+1 = ¹⁰⁰C_r . 5^{(100 – r)/2} n^r/4

Herein, r = 0, 1, 2, . . . . . . , 100

r must be 0, 4, 8, … 100

Thus, it can be said:

Number of rational terms = 26

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Long Answer Questions [3 Marks Questions]

Ques. Find the coefficient of x⁵ in binomial expansion of (1 + 2x)⁵ (1 - x)⁷.

Ans. Using the binomial theorem we will expand both terms.

We know that, 

(x+y)ⁿ = ⁿC₀xⁿ + ⁿC₁x^n-1y + ⁿC₂x^n-2y² + ⁿC₃x^n-3y³ +........+ ⁿC_rx^n-ry^r +..... + ⁿC₀x⁰yⁿ

Applying the formula we get,

(1 + 2x)⁵ (1 - x)⁷ = (1 + ⁶C₁(2x) + ⁶C₂(2x)² + ⁶C₃(2x)³ + ⁶C₄(2x)⁴ + ⁶C₅(2x)⁵ + ⁶C₆(2x)₆) 

(1 - ⁷C₁x + ⁷C₂(x)² + ⁷C₃(x)³ + ⁷C₄(x)₄ + ⁷C₅(x)⁵ + ⁷C₆(x)⁶ + ⁷C₇(x)⁷)

= (1 + 12x + 60x² + 160x³ + 240x⁴ + 192x⁵ + 64x⁶) (1 - 7x + 21x² - 35x³ + 35x⁴ - 21x⁵ + 7x⁶ - x⁷)

Clearly, it can be determined that the coefficient of x⁵ is 

⇒ 1 * (-21) + 12 * 35 + 60(-35) + 160 * 21 + 240 * (-7) + 192 * 1

⇒ 171

Therefore, the coefficient of x5 in (1 + 2x)⁵ (1 - x)⁷ is 171.

Ques. What are the last three digits of 27²⁶?

Ans. in order to determine the same, the value 27²⁶ needs to be reduced into the form (730 – 1)ⁿ. After using simple binomial expansion, the digits can be obtained.

Now, we have, 27² = 729

Thus,

27²⁶ = (729)¹³ = (730 – 1)¹³

= ¹³C₀ (730)¹³ – ¹³C₁ (730)¹² + ¹³C₂ (730)¹¹ – . . . . . – ¹³C₁₀ (730)³ + ¹³C₁₁(730)² – ¹³C₁₂ (730) + 1

= 1000m + [(13 × 12)]/2] × (14)² – (13) × (730) + 1

Herein, we can say that ‘m’ is a positive integer

Thus,

= 1000m + 15288 – 9490 = 1000m + 5799

Therefore, the last three digits of 17²⁵⁶ are 799.

Ques. Calculate the value of (101)⁴ by using the binomial theorem.

Ans. As per the given value, (101)^4.

As can be seen, here, 101 can also be written as the sum or the difference of two numbers, in a way that the binomial theorem can be used.

Thus, 101 = 100+1

Hence, (101)⁴ = (100+1)⁴

Now, after the binomial theorem is applied, we get:

(101)⁴ = (100+1)⁴ = ⁴C₀(100)⁴ +⁴C₁ (100)³(1) + ⁴C₂(100)²(1)² +⁴C₃(100)(1)³ +⁴C₄(1)⁴

→ (101)⁴ = (100)⁴+4(100)³+6(100)²+4(100) + (1)⁴

→ (101)⁴ = 100000000+ 4000000+ 60000+ 400+1

→ (101)⁴ = 104060401

Hence, the value of (101)⁴ is 104060401.

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Very Long Questions [5 Marks Questions]

Ques. By applying the binomial theorem, represent that 6ⁿ – 5n always leaves behind remainder 1 after it is divided by 25

Ans. Consider that for any two given numbers, assume x and y, the numbers q and r can be determined such that x = yq + r. After that, it can be said that b divides x with q as the quotient and r as the remainder. Thus, in order to illustrate that 6ⁿ – 5n leaves the remainder 1 after being divided by 25, we should also prove that 6ⁿ – 5n = 25k + 1, (here, k is a natural number).

Now, we are aware that, (1 + a)ⁿ = ⁿC₀ + ⁿC₁ a + ⁿC₂ a² + … + ⁿC_n aⁿ

Now for a = 5, we can obtain the following:

(1 + 5)ⁿ = ⁿC₀ + ⁿC₁ 5 + ⁿC₂ (5)² + … + ⁿC_n 5ⁿ

Thus, the above form can also be represented as:

6ⁿ = 1 + 5n + 5^2 nC₂ + 5^3 nC₃+ ….+ 5ⁿ

Now, get 5n to the L.H.S, thus obtaining,

→ 6ⁿ – 5n = 1 + 5^2 nC₂ + 5^3 nC₃+ ….+ 5ⁿ

→ 6ⁿ – 5n = 1 + 5² (ⁿC₂ + 5 ⁿC₃+ ….+ 5^n-2)

→ 6ⁿ – 5n = 1 + 25 (ⁿC₂ + 5 ⁿC₃+ ….+ 5^n-2)

→ 6ⁿ – 5n = 1 + 25 k (here, k =ⁿC₂ + 5 ⁿC₃+ ….+ 5^n-2)

Hence, the above form proves that, when 6ⁿ–5n is divided by 25, leaving the remainder 1.

Hence, the given statement is proven.

Ques. Solve the following:

1. Determine the degree of the polynomial [x + {√(3^(3-1))}^1/2]⁵ + [x + {√(3^(3-1))}^1/2]⁵.
2. Illustrate that 11⁹ + 9¹¹ is divisible by 10.
3. Determine the larger value of 99⁵⁰ + 100⁵⁰ and 101⁵⁰

4. Assuming that the coefficients of three consecutive terms in the expansion of (1+x)ⁿ are in the ratio 1:7:42. Hence, determine the value of n.

Ans. a) As per the question, it can be said, [x + { √(3^(3-1)) }^1/2 ]⁵:

= 2 [⁵C₀ x⁵ + ⁵C₂ x⁵ (x³ – 1) + ⁵C₄ . x . (x³ – 1)²]

Thus, the highest power = 7.

​b) It can be showed that: 11⁹ + 9¹¹ = (10 + 1)⁹ + (10 − 1)¹¹

= (9C₀ . 10⁹ + 9C₁ . 10⁸ + … 9C₉) + (11C₀ . 10¹¹ − 11C₁ . 10¹⁰ + … −11C₁₁)

= 9C₀ . 10⁹ + 9C₁ . 10⁸ + … + 9C₈ . 10 + 1 + 10¹¹ − 11C₁ . 10¹⁰ + … + 11C₁₀ . 10−1

= 10[9C₀ . 10⁸ + 9C₁ . 10⁷ + … + 9C₈ + 11C₀ . 10¹⁰ − 11C₁ . 10⁹ + … + 11C₁₀]

= 10K, (Thus, it is divisible by 10).​

c) 101⁵⁰ = (100 + 1)⁵⁰ = 100⁵⁰ + 50 . 100⁴⁹ + 25.49 . 100⁴⁸ + …

⇒ 99⁵⁰ = (100 − 1)⁵⁰ = 100⁵⁰ – 50 . 100⁴⁹ + 25.49 . 100⁴⁸ − ….

⇒ 101⁵⁰ – 99⁵⁰ = 2[50 . 100⁴⁹ + 25(49) (16) 100⁴⁷ + …]

= 100⁵⁰ + 50 . 49 . 16 . 100⁴⁷ + … >100⁵⁰

∴ 101⁵⁰ – 99⁵⁰ > 100⁵⁰

⇒ 101⁵⁰ > 100⁵⁰ + 99⁵⁰

d) Consider (r – 1)^th, (r)^th and (r + 1)^th are three consecutive terms.

Thus, the given ratio can be given by:1:7:42

Hence, (ⁿC_r-2 / ⁿC_{r – 1}) = (1/7)

(ⁿC_r-2 / ⁿC_{r – 1}) = (1/7) ⇒ [(r – 1)/(n − r+2)] = (1/7) ⇒ n−8r+9=0 → (1)

And,

(ⁿC_r-1 / ⁿC_r) = (7/42) ⇒ [(r)/(n – r +1)] =(1/6) ⇒ n−7r +1=0 → (2)

From (1) & (2), n = 55

Ques. What are the properties or Binomial Theorem? If (1 + x)¹⁵ = a₀ + a₁x + . . . . .  + a₁₅ x¹⁵ then, determine the value of .

Ans. The many properties of Binomial Theorem are:

• C₀ + C₁ + C₂ + … + C_n = 2ⁿ
• C₀ + C₂ + C₄ + … = C₁ + C₃ + C₅ + … = 2^n-1
• C₀ – C₁ + C₂ – C₃ + … +(−1)ⁿ . nC_n = 0
• nC₁ + 2.nC₂ + 3.nC₃ + … + n.nC_n = n.2^n-1
• C₁ − 2C₂ + 3C₃ − 4C₄ + … +(−1)^n-1 C_n = 0 for n > 1
• C₀² + C₁² + C₂² + …C_n² = [(2n)!/ (n!)²]

as for the question given,

Thus, it can be shown,

= C₁/C₀ + 2 C₂/C₁+ 3C₃/C₂ + . . . . + 15 C₁₅/C₁₄

= 15 + 14 + 13 + . . . . . + 1 = [15(15+1)]/2 = 120

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