Chapter 4 Methods of Induction and Binomial Theorem Miscellaneous Exercise 4 Solutions

Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Miscellaneous Exercise 4

(I) Select the correct answers from the given alternatives.

(II) Answer the following.

Step II:
Let us assume that P(n) is true for n = k.
∴ 8 + 17 + 26 +…..+ (9k – 1) = (9k + 7) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., 8 + 17 + 26 + …… + [9(k + 1) – 1]

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 8 + 17 + 26 +…..+ (9n – 1) = (9n + 7) for all n ∈ N.

(ii) 1²+ 4²+ 7²+ …… + (3n – 2)²= (6n²– 3n – 1)
Solution:
Let P(n) = 1²+ 4²+ 7²+ ….. + (3n – 2)²= (6n²– 3n – 1), for all n ∈ N.
Step I:
Put n = 1
L.H.S.= 1²= 1
R.H.S.= [6(1)²– 3(1) – 1] = 1
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 1²+ 4²+ 7²+…..+ (3k – 2)²= (6k²– 3k – 1) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1²+ 4²+ 7²+ … + (3n – 2)²= (6n²– 3n – 1) for all n ∈ N.

(iii) 2 + 3.2 + 4.2²+ …… + (n + 1) 2^n-1= n. 2ⁿ
Solution:
Let P(n) ≡ 2 + 3.2 + 4.2²+…..+ (n + 1) 2^n-1= n.2ⁿ, for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 2
R.H.S. = 1(2¹) = 2
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 2 + 3.2 + 4.2²+ ….. + (k + 1) 2^k-1= k.2^k…..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
2 + 3.2 + 4.2²+….+ (k + 2) 2^k= (k + 1) 2^k+1

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 2 + 3.2 + 4.2²+……+ (n + 1) 2^n-1= n.2ⁿfor all n ∈ N.

(iv) =
Solution:

Step II:
Let us assume that P(n) is true for n = k.
∴ t_k+1= 5t_k– 8 and t_k= 5^k-1+ 2

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
t_k+1= 5^k+1-1+ 2 = 5^k+ 2
t_k+1= 5t_k– 8 and t_k= 5^k-1+ 2 ……[From Step II]
∴ t_k+1= 5(5^k-1+ 2) – 8 = 5^k+ 2
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ t_n= 5^n-1+ 2, for all n ∈ N.

(ii)
Solution:
Here, a = x, b = , n = 10.
Now, n is even.
∴
∴ Middle term is t6, for which r = 5

(iii) (x²+ 2y²)⁷
Solution:
Here, a = x², b = 2y², n = 7.
Now, n is odd.
∴
∴ Middle terms are t₄and t₅, for which r = 3 and r = 4 respectively.

∴ Middle terms are 280x⁸y⁶and 560x⁶y⁸.

(iv)
Solution:

(ii) x⁶⁰in the expansion of
Solution:

(ii)
Solution:

(ii) 15^2n-1+ 1 is divisible by 16, for all n ∈ N.
Solution:
15^2n-1+ 1 is divisible by 16, if and only if (15^2n-1+ 1) is is a multiple of 16.
Let P(n) ≡ 15^2n-1+ 1 = 16m, where m ∈ N.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 15^2n-1+ 1 is divisible by 16, for all n ∈ N.

(iii) 5²ⁿ– 2²ⁿis divisible by 3, for all n ∈ N.
Solution: