NCERT Solutions for Class 11 Maths Chapter 4

A meaningful sentence which can be judged to be either true or false is called a statement.

A statement involving mathematical relations is called as mathematical statement.

Principle of mathematical Induction:
1. Let P(n) be any statement involving natural number n such that
P(1) is true, and
2. If P(k) is true as well as P(k + 1) is true for some natural number k, i.e. P(k + 1) is true whenever P(k) is true for some natural number k, then P(n) is true for all natural numbers.

Let the given statement be P(n), therefore,
P(n):1.2+〖2.2〗^2+〖3.2〗^3+⋯+n.2^n = (n-1) 2^(n+1)+2
For n = 1, we have L.H.S. = 1.2 = 2 R.H.S. = (1-1) 2^(1+1)+2 = 0+2=2
L.H.S. = R.H.S, so P(1) is true.
Let P(k) be true for some positive integer k, such that
P(k):1.2+〖2.2〗^2+〖3.2〗^3+⋯+k.2^k = (k-1) 2^(k+1)+2
Now, to prove that P(k + 1) is true. i.e. P(k+1):1.2+〖2.2〗^2+〖3.2〗^3+⋯+k.2^k+(k+1).2^(k+1) = (k) 2^(k+2)+2
Considering the L.H.S. of P(k + 1), we have 1.2+〖2.2〗^2+〖3.2〗^3+⋯+k.2^k+(k+1).2^(k+1)
= (1.2+〖2.2〗^2+〖3.2〗^3+⋯+k.2^k )+(k+1).2^(k+1)
= (k-1) 2^(k+1)+2+(k+1).2^(k+1) [As P(k) is true]
= (k-1) 2^(k+1)+(k+1).2^(k+1)+2
= [(k-1)+(k+1)].2^(k+1)+2 =2k.2^(k+1)+2
= k.2^(k+2)+2 =R.H.S.
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the Principle of Mathematical Induction, statement P(n) is true for all natural numbers.

Let the given statement be P(n), therefore, P(n):〖10〗^(2n-1)+1 is divisible by 11.
For n = 1, we have 〖10〗^(2-1)+1=11,which is divisible by 11.
So, P(1) is true.
Let P(k) be true for some positive integer k, such that P(k):〖10〗^(2k-1)+1 is divisible by 11.
Let 〖10〗^(2n-1)+1 = 11m … (1) Where, m is any natural number.
Now, to prove that P(k + 1) is true. i.e. P(k+1):〖10〗^(2k+1)+1 is divisible by 11.
Consider 〖10〗^(2k+1)+1 =〖10〗^(2k-1+2)+1
=〖10〗^2.〖10〗^(2k-1)+1
=〖10〗^2.(11m-1)+1 [From the equation (1),〖10〗^(2n-1) = 11m-1]
= 100.(11m-1)+1 =1100m – 100 + 1
= 1100m – 99
= 11[100m – 9], which is divisible by 11.
Thus, P(k + 1) is true whenever P(k) is true. Hence, by the Principle of Mathematical Induction, statement P(n) is true for all natural numbers.

Let the given statement be P(n), therefore, P(n):x²n-y²n is divisible by x+y.
For n = 1, we have x²-y²=(x+y)(x-y), which is divisible by x+y.
So, P(1) is true.
Let P(k) be true for some positive integer k, such that P(k):x²k-y²k is divisible by x+y.
Let x²k-y²k=(x+y)m … (1) Where, m is any natural number.
Now, to prove that P(k + 1) is true. i.e. P(k+1):x²(k+1) – y²(k+1) is divisible by x+y.
Consider x^2(k+1) -y^2(k+1)
= x^(2k+2)-y^(2k+2) =〖x^2 x〗^2k-y^(2k+2)
= x² [(x+y)m+y^2k ]-y^(2k+2) [From the equation (1),x^2k=(x+y)m+y^2k ]
= x² (x+y)m+x^2 y^2k-y^(2k+2)
= x² (x+y)m+y^2k (x²-y²)
= x² (x+y)m+y^2k (x-y)(x+y) =(x+y)[x^2 m+y^2k (x-y)], which is divisible by (x+y).
Thus, P(k + 1) is true whenever P(k) is true. Hence, by the Principle of Mathematical Induction, statement P(n) is true for all natural numbers.

Let the given statement be P(n), therefore, P(n):(2n+7)<(n+3)². For n = 1, we have (2×1+7)<(1+3)² ⇒ 9 < 16 ⇒ P(1) is true. Let P(k) be true for some positive integer k, such that P(k):(2k+7)<(k+3)². Now, to prove that P(k + 1) is true. i.e. P(k+1):(2k+9)<(k+4)² Considering the statement P(k), we have (2k+7)<(k+3)² [Difference between (k+4)² and (k+3)² = (k+4)² - (k+3)² = k²+8k+16-(k²+6k+9)=2k+7] Adding both sides (2k+7), we have (2k+7) + (2k+7)<(k+3)² + (2k+7) ⇒(2k+9) + (2k+5) < (k² +6k + 9) + (2k+7) ⇒(2k+9)+(2k+5) < (k² + 6k + 9 + 2k + 7) ⇒(2k+9)+(2k+5) < (k² +8k + 16) ⇒(2k+9)+(2k+5) < (k+4)² ⇒(2k+9)<(k+4)² [As k>0,so 2k+1>0]
Thus, P(k + 1) is true whenever P(k) is true. Hence, by the Principle of Mathematical Induction, statement P(n) is true for all natural numbers.