Important Questions: Binomial Theorem | Mathematics (Maths) Class 11 - Commerce PDF Download

Q1: Evaluate (101)⁴ using the binomial theorem.
Ans: Given: (101)⁴.
Here, 101 can be written as the sum or the difference of two numbers, such that the binomial theorem can be applied.
Therefore, 101 = 100+1
Hence, (101)⁴ = (100+1)⁴
Now, by applying the binomial theorem, 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.

Q2: Find the value of r, If the coefficients of (r – 5)th and (2r – 1)th terms in the expansion of (1 + x)³⁴ are equal.
Ans: For the given condition, the coefficients of (r – 5)th and (2r – 1)th terms of the expansion (1 + x)³⁴ are ³⁴C_r-6 and ³⁴C_2r-2 respectively.
Since the given terms in the expansion are equal,
³⁴C_r-6 = ³⁴C_2r-2
From this, we can write it as either
r-6=2r-2
(or)
r-6=34 -(2r-2) [We know that, if nCr = nCp , then either r = p or r = n – p
So, we get either r = – 4 or r = 14.
We know that r being a natural number, the value of r = – 4 is not possible.
Hence, the value of r is 14.

Q3: Expand the expression (2x-3)6 using the binomial theorem.
Ans: Given Expression: (2x-3)6
By using the binomial theorem, the expression (2x-3)6 can be expanded as follows:
(2x-3)⁶ = ⁶C₀(2x)⁶ –⁶C₁(2x)⁵(3) + ⁶C₂(2x)⁴(3)² – ⁶C₃(2x)³(3)³ + ⁶C₄(2x)²(3)⁴ – ⁶C₅(2x)(3)⁵ + ⁶C₆(3)⁶
(2x-3)⁶ = 64x⁶ – 6(32x⁵ )(3) +15(16x⁴ )(9) – 20(8x³ )(27) +15(4x² )(81) – 6(2x)(243) + 729
(2x-3)⁶ = 64x⁶ -576x⁵ + 2160x⁴ – 4320x³ + 4860x² – 2916x + 729
Thus, the binomial expansion for the given expression (2x-3)⁶ is 64x⁶ -576x⁵ + 2160x⁴ – 4320x³ + 4860x² – 2916x + 729.

Q4: Using the binomial theorem, show that 6ⁿ–5n always leaves remainder 1 when divided by 25
Ans: Assume that, for any two numbers, say x and y, we can find numbers q and r such that x = yq + r, then we say that b divides x with q as quotient and r as remainder. Thus, in order to show that 6n – 5n leaves remainder 1 when divided by 25, we should prove that 6ⁿ – 5n = 25k + 1, where k is some natural number.
We know that,
(1 + a)ⁿ = ⁿC₀ + ⁿC₁ a + ⁿC₂ a² + … + ⁿC_n aⁿ
Now for a=5, we get:
(1 + 5)ⁿ = ⁿC₀ + ⁿC₁ 5 + ⁿC₂ (5)² + … + ⁿC_n 5ⁿ
Now the above form can be written as:
6ⁿ = 1 + 5n + 5^{2 n}C² + 53 ⁿC₃ + ….+ 5ⁿ
Now, bring 5n to the L.H.S, we get
6ⁿ – 5n = 1 + 5^{2 n}C₂ + 53 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 (where k = ⁿC₂ + 5 ⁿC₃ + ….+ 5^n-2)
The above form proves that, when 6n–5n is divided by 25, it leaves the remainder 1.
Hence, the given statement is proved.