Class 11 NCERT Solutions Chapter 4 Principle of Mathematical Induction Exercise 4.1 | Set 1 (original) (raw)
Last Updated : 23 Jul, 2025
The Principle of Mathematical Induction is a fundamental concept in mathematics used to the prove statements or formulas that are asserted for the every natural number. This principle is pivotal in the establishing the validity of the various mathematical propositions and theorems.
Principle of Mathematical Induction
The Principle of Mathematical Induction involves the three main steps:
- Base Case: The Verify the statement for the initial value usually n=1.
- Inductive Step: The Assume the statement is true for the n=k.
- Inductive Proof: Prove that the statement is true for the n=k+1 based on the Inductive Hypothesis.
Prove the following by using the principle of mathematical induction for all n ∈ N:
Question 1: 1 + 3 + 32 + ........ + 3n-1 = \frac{3^n - 1}{2}
**Solution:
We have,
P(n) = 1 + 3 + 32 + ........ + 3n-1 = \frac{3^n - 1}{2}
For **n=1, we get
P(1) = 1 = \frac{3^1 - 1}{2} = \frac{3-1}{2} = 1
So, P(1) is true
Assume that P(k) is true for some positive integer **n=k
P(k) = 1 + 3 + 32 + ........ + 3k-1 = \frac{3^k - 1}{2} .................(1)
Let's prove that P(k + 1) is also true. Now, we have
**P(k+1) = 1 + 3 + 32 + ........ + 3k-1 + 3(k+1)-1
= (1 + 3 + 32 + ........ + 3k-1) + 3k
From Eq(1), we get
= \frac{3^k - 1}{2} + 3k
= \frac{3^k - 1 + 2(3^k)}{2}
= \frac{3.3^k - 1}{2}
= \frac{3^{k+1} - 1}{2}
Hence,
P(k+1) = \frac{3^{k+1} - 1}{2}
**Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is **true for all natural numbers n.
Question 2: 1 + 23 + 33 + ........... + n3 = [\frac{n(n+1)}{2} ]^2
**Solution:
We have,
P(n) = 1 + 23 + 33 + ........... + n3 = [\frac{n(n+1)}{2} ]^2
For **n=1, we get
P(1) = 1 = [\frac{1(1+1)}{2} ]^2 = 1
So, P(1) is true
Assume that P(k) is true for some positive integer **n=k
P(k) = 1 + 23 + 33 + ........... + k3 = [\frac{k(k+1)}{2} ]^2 .................(1)
Let's prove that P(k + 1) is also true. Now, we have
**P(k+1) = 1 + 23 + 33 + ........... + k3 + (k+1)3
= (1 + 23 + 33 + ........... + k3) + (k+1)3
From Eq(1), we get
= [\frac{k(k+1)}{2} ]^2 + (k+1)3
= \frac{k^2(k+1)^2}{4} + (k+1)3
= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}
Taking (k+1)2, we get
= \frac{(k+1)^2(k^2 + 4(k+1))}{4}
= \frac{(k+1)^2(k^2 + 4k + 4)}{4}
= \frac{(k+1)^2(k+2)^2}{2^2}
= [\frac{(k+1)((k+1)+1)}{2} ]^2
Hence,
P(k+1) = [\frac{(k+1)((k+1)+1)}{2} ]^2
**Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is **true for all natural numbers n.
Question 3: 1 + \frac{1}{(1+2)} + \frac{1}{(1+2+3)} + ....... + \frac{1}{(1+2+3+....n)} = \frac{2n}{(n+1)}
**Solution:
We have,
P(n) = 1 + \frac{1}{(1+2)} + \frac{1}{(1+2+3)} + ....... + \frac{1}{(1+2+3+....n)} = \frac{2n}{(n+1)}
For **n=1, we get
P(1) = 1 = \frac{2(1)}{1+1} = 1
So, P(1) is true
Assume that P(k) is true for some positive integer **n=k
P(k) = 1 + \frac{1}{(1+2)} + \frac{1}{(1+2+3)} + ....... + \frac{1}{(1+2+3+....k)} = \frac{2k}{(k+1)} .................(1)
Let's prove that P(k + 1) is also true. Now, we have
**P(k+1) = 1 + \frac{1}{(1+2)} + \frac{1}{(1+2+3)} + ....... + \frac{1}{(1+2+3+....k)} + \frac{1}{(1+2+3+....k+(k+1))}
= (1 + \frac{1}{(1+2)} + \frac{1}{(1+2+3)} + ....... + \frac{1}{(1+2+3+....k)} ) + \frac{1}{(1+2+3+....k+(k+1))}
From Eq(1), we get
= \frac{2k}{(k+1)} + \frac{1}{(1+2+3+....k+(k+1))}
As we know that,
Sum of first natural number,
1 + 2 + 3 + ...... + n = \frac{n(n+1)}{2}
So, we get
= \frac{2k}{(k+1)} + \frac{1}{\frac{(k+1)(k+1+1)}{2}}
= \frac{2k}{(k+1)} + \frac{2}{(k+1)(k+2)}
= \frac{2}{(k+1)} (k+\frac{1}{k+2})
= \frac{2}{(k+1)} (\frac{(k(k+2) + 1)}{k+2})
= \frac{2}{(k+1)} (\frac{(k^2+2k+1)}{k+2})
= \frac{2}{(k+1)} (\frac{(k+1)^2}{k+2})
= \frac{2(k+1)}{(k+1)+1}
Hence,
P(k+1) = \frac{2(k+1)}{(k+1)+1}
**Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is **true for all natural numbers n.
Question 4: 1.2.3 + 2.3.4 +…+ n(n+1) (n+2) = \frac{n(n+1)(n+2)(n+3)}{4}
**Solution:
We have,
P(n) = 1.2.3 + 2.3.4 +…+ n(n+1) (n+2) = \frac{n(n+1)(n+2)(n+3)}{4}
For **n=1, we get
P(1) = 1.2.3 = \frac{1(1+1)(1+2)(1+3)}{4} = \frac{1.2.3.4}{4} = 1.2.3
So, P(1) is true
Assume that P(k) is true for some positive integer **n=k
P(k) = 1.2.3 + 2.3.4 +…+ k(k+1) (k+2) = \frac{k(k+1)(k+2)(k+3)}{4} .................(1)
Let's prove that P(k + 1) is also true. Now, we have
**P(k+1) = 1.2.3 + 2.3.4 +…+ k(k+1) (k+2) + (k+1)(k+1+1) (k+1+2)
= (1.2.3 + 2.3.4 +…+ k(k+1) (k+2)) + (k+1)(k+2) (k+3)
From Eq(1), we get
= \frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3)
= (k+1)(k+2) (k+3) (\frac{k}{4} + 1)
= (k+1)(k+2) (k+3) (\frac{k + 4}{4})
= \frac{(k+1)((k+1)+1) ((k+1)+2) ((k+1)+3)}{4}
Hence,
P(k+1) = \frac{(k+1)((k+1)+1) ((k+1)+2) ((k+1)+3)}{4}
**Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.
Question 5: 1.3 + 2.32 + 3.33 +…+ n.3n = \frac{(2n-1)3^{n+1}+3}{4}
**Solution:
We have,
P(n) = 1.3 + 2.32 + 3.33 +…+ n.3n = \frac{(2n-1)3^{n+1}+3}{4}
For **n=1, we get
P(1) = 1.3 = 3 = \frac{(2(1)-1)3^{1+1}+3}{4} = \frac{9+3}{4} = \frac{12}{4} = 3
So, P(1) is true
Assume that P(k) is true for some positive integer **n=k
P(k) = 1.3 + 2.32 + 3.33 +…+ k.3k = \frac{(2k-1)3^{k+1}+3}{4} .................(1)
Let's prove that P(k + 1) is also true. Now, we have
**P(k+1) = 1.3 + 2.32 + 3.33 +…+ k.3k + (k+1).3(k+1)
= (1.3 + 2.32 + 3.33 +…+ k.3k) + (k+1).3(k+1)
From Eq(1), we get
= \frac{(2k-1)3^{k+1}+3}{4} + (k+1).3(k+1)
= \frac{(2k-1)3^{k+1}+3 + 4(k+1).3^{k+1}}{4}
= 3k+1 \frac{((2k-1) + 4(k+1)) + 3}{4}
= 3k+1 \frac{(6k+3) + 3}{4}
= 3k+1 \frac{(3(2k+1)) + 3}{4}
= 3(k+1)+1 \frac{(2(k+1)-1) + 3}{4}
Hence,
P(k+1) = 3(k+1)+1 \frac{(2(k+1)-1) + 3}{4}
**Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is **true for all natural numbers n.
Question 6: 1.2 + 2.3 + 3.4 +…+ n.(n+1) = [\frac{n(n+1)(n+2)}{3}]
**Solution:
We have,
P(n) = 1.2 + 2.3 + 3.4 +…+ n.(n+1) = [\frac{n(n+1)(n+2)}{3}]
For **n=1, we get
P(1) = 1.2 = 2 = [\frac{1(1+1)(1+2)}{3}] = [\frac{1(2)(3)}{3}] = 2
So, P(1) is true
Assume that P(k) is true for some positive integer **n=k
P(k) = 1.2 + 2.3 + 3.4 +…+ k.(k+1) = [\frac{k(k+1)(k+2)}{3}] .................(1)
Let's prove that P(k + 1) is also true. Now, we have
**P(k+1) = 1.2 + 2.3 + 3.4 +…+ k.(k+1) + (k+1)(k+1+1)
= (1.2 + 2.3 + 3.4 +…+ k.(k+1)) + (k+1)(k+2)
From Eq(1), we get
= [\frac{k(k+1)(k+2)}{3}] + (k+1)(k+2)
= [\frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}]
= (k+1)(k+2) [\frac{k+3}{3}]
= [\frac{(k+1)(k+2)(k+3)}{3}]
= [\frac{(k+1)((k+1)+1)((k+1)+2)}{3}]
Hence,
P(k+1) = [\frac{(k+1)((k+1)+1)((k+1)+2)}{3}]
**Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is **true for all natural numbers n.
Question 7: 1.3 + 3.5 + 5.7 +…+ (2n–1) (2n+1) = \frac{n(4n^2+6n-1)}{3}
**Solution:
We have,
P(n) = 1.3 + 3.5 + 5.7 +…+ (2n–1) (2n+1) = \frac{n(4n^2+6n-1)}{3}
For **n=1, we get
P(1) = 1.3 = 3 = \frac{1(4(1)^2+6(1)-1)}{3} = \frac{9}{3} = 3
So, P(1) is true
Assume that P(k) is true for some positive integer **n=k
P(k) = 1.3 + 3.5 + 5.7 +…+ (2k–1) (2k+1) = \frac{k(4k^2+6k-1)}{3} .................(1)
Let's prove that P(k + 1) is also true. Now, we have
**P(k+1) = 1.3 + 3.5 + 5.7 +…+ (2k–1) (2k+1) + (2(k+1)-1)(2(k+1)+1)
= (1.3 + 3.5 + 5.7 +…+ (2k–1) (2k+1)) + (2k+1)(2k+3)
From Eq(1), we get
= \frac{k(4k^2+6k-1)}{3} + (4k2+8k+3)
= \frac{k(4k^2+6k-1)+3(4k^2+8k+3)}{3}
= \frac{4k^3+6k^2-k+12k^2+24k+9}{3}
= \frac{4k^3+18k^2+23k+9}{3}
= \frac{4k^3+14k^2+9k+4k^2+14k+9}{3}
= \frac{k(4k^2+14k+9)+1(4k^2+14k+9)}{3}
= \frac{(k+1) (4k^2+14k+9)}{3}
= \frac{(k+1) (4(k^2+2k+1)+6(k+1)-1)}{3}
= \frac{(k+1) (4(k+1)^2+6(k+1)-1)}{3}
Hence,
P(k+1) = \frac{(k+1) (4(k+1)^2+6(k+1)-1)}{3}
**Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is **true for all natural numbers n.
Question 8: 1.2 + 2.22 + 3.23 + ...+n.2n = (n–1) 2n + 1 + 2
**Solution:
We have,
P(n) = 1.2 + 2.22 + 3.23 + ...+n.2n = (n–1) 2n + 1 + 2
For **n=1, we get
P(1) = 1.2 = 2 = (1–1) 2(1) + 1 + 2 = 2
So, P(1) is true
Assume that P(k) is true for some positive integer **n=k
P(k) = 1.2 + 2.22 + 3.23 + ...+k.2k = (k–1) 2k + 1 + 2 .................(1)
Let's prove that P(k + 1) is also true. Now, we have
**P(k+1) = 1.2 + 2.22 + 3.23 + ...+k.2k + (k+1).2(k+1)
= (1.2 + 2.22 + 3.23 + ...+k.2k) + (k+1).2(k+1)
From Eq(1), we get
= (k–1) 2k + 1 + 2 + (k+1).2k+1
= 2k + 1((k–1) + (k+1)) + 2
= 2k + 1(2k) + 2
= k.2k+1+1 + 2
= ((k+1)-1).2(k+1)+1 + 2
Hence,
P(k+1) = ((k+1)-1).2(k+1)+1 + 2
**Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.
Question 9: \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...... + \frac{1}{2^n} = 1 - \frac{1}{2^n}
**Solution:
We have,
P(n) = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...... + \frac{1}{2^n} = 1 - \frac{1}{2^n}
For **n=1, we get
P(1) = \frac{1}{2} = 1 - \frac{1}{2^1} = \frac{1}{2}
So, P(1) is true
Assume that P(k) is true for some positive integer **n=k
P(k) = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...... + \frac{1}{2^k} = 1 - \frac{1}{2^k} .................(1)
Let's prove that P(k + 1) is also true. Now, we have
**P(k+1) = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...... + \frac{1}{2^k} + \frac{1}{2^{k+1}}
= (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...... + \frac{1}{2^k} ) + \frac{1}{2^{k+1}}
From Eq(1), we get
= 1 - \frac{1}{2^k} + \frac{1}{2^{k+1}}
= 1 - \frac{1}{2^k} + \frac{1}{2.2^k}
= 1 - \frac{1}{2^k}(1-\frac{1}{2})
= 1 - \frac{1}{2^k}(\frac{1}{2})
= 1 - \frac{1}{2^{k+1}}
Hence,
P(k+1) = 1 - \frac{1}{2^{k+1}}
**Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.
Question 10: \frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + ...... + \frac{1}{(3n-1)(3n+2)} = \frac{n}{(6n+4)}
**Solution:
We have,
P(n) = \frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + ...... + \frac{1}{(3n-1)(3n+2)} = \frac{n}{(6n+4)}
For **n=1, we get
P(1) = \frac{1}{2.5} = \frac{1}{10} = \frac{1}{(6(1)+4)} = \frac{1}{10}
So, P(1) is true
Assume that P(k) is true for some positive integer **n=k
P(k) = \frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + ...... + \frac{1}{(3k-1)(3k+2)} = \frac{k}{(6k+4)} .................(1)
Let's prove that P(k + 1) is also true. Now, we have
**P(k+1) = \frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + ...... + \frac{1}{(3k-1)(3k+2)} + \frac{1}{(3(k+1)-1)(3(k+1)+2)}
= (\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + ...... + \frac{1}{(3k-1)(3k+2)} ) + \frac{1}{(3k+2)(3k+5)}
From Eq(1), we get
= \frac{k}{(6k+4)} + \frac{1}{(3k+2)(3k+5)}
= \frac{k}{2(3k+2)} + \frac{1}{(3k+2)(3k+5)}
= \frac{1}{3k+2} (\frac{k}{2} + \frac{1}{3k+5})
= \frac{1}{3k+2} (\frac{k(3k+5)+1(2)}{2(3k+5)})
= \frac{1}{3k+2} (\frac{3k^2+5k+2}{6k+10})
= \frac{1}{3k+2} (\frac{(3k+2)(k+1)}{6k+10})
= \frac{(k+1)}{6(k+1)+4}
Hence,
P(k+1) = \frac{(k+1)}{6(k+1)+4}
**Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is **true for all natural numbers n.
Question 11: \frac{1}{1.2.3} + \frac{1}{2.3.5} + \frac{1}{3.4.5} + ...... + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}
**Solution:
We have,
P(n) = \frac{1}{1.2.3} + \frac{1}{2.3.5} + \frac{1}{3.4.5}+ ...... + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}
For **n=1, we get
P(1) = \frac{1}{1.2.3} = \frac{1}{6} = \frac{1(1+3)}{4(1+1)(1+2)} = \frac{4}{4(2)(3)} = \frac{1}{6}
So, P(1) is true
Assume that P(k) is true for some positive integer **n=k
P(k) = \frac{1}{1.2.3} + \frac{1}{2.3.5} + \frac{1}{3.4.5}+ ...... + \frac{1}{k(k+1)(k+2)} = \frac{k(k+3)}{4(k+1)(k+2)} .................(1)
Let's prove that P(k + 1) is also true. Now, we have
**P(k+1) = \frac{1}{1.2.3} + \frac{1}{2.3.5} + \frac{1}{3.4.5}+ ...... + \frac{1}{k(k+1)(k+2)} + \frac{1}{(k+1)(k+1+1)(k+1+2)}
= (\frac{1}{1.2.3} + \frac{1}{2.3.5} + \frac{1}{3.4.5}+ ...... + \frac{1}{k(k+1)(k+2)} ) + \frac{1}{(k+1)(k+2)(k+3)}
From Eq(1), we get
= \frac{1}{(k+1)(k+2)} (\frac{k(k+3)}{4} + \frac{1}{(k+3)})
= \frac{1}{(k+1)(k+2)} (\frac{k(k+3)^2 + 4}{4(k+3)})
= \frac{1}{(k+1)(k+2)} (\frac{k(k^2+6k+9) + 4}{4(k+3)})
= \frac{1}{(k+1)(k+2)} (\frac{k^3+6k^2+9k + 4}{4(k+3)})
= \frac{1}{(k+1)(k+2)} (\frac{k^3+2k^2+k + 4k^2 + 8k+ 4}{4(k+3)})
= \frac{1}{(k+1)(k+2)} (\frac{k(k^2+2k+1) + 4(k^2 + 2k+ 1)}{4(k+3)})
= \frac{1}{(k+1)(k+2)} (\frac{(k+4)(k^2+2k+1)}{4(k+3)})
= \frac{1}{(k+1)(k+2)} (\frac{(k+4)(k+1)^2}{4(k+3)})
= \frac{1}{(k+2)} (\frac{(k+4)(k+1)}{4(k+3)})
= \frac{(k+1)((k+1)+3)}{4((k+1)+1)((k+1)+2))}
Hence,
P(k+1) = \frac{(k+1)((k+1)+3)}{4((k+1)+1)((k+1)+2))}
**Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is **true for all natural numbers n.
Question 12: a + ar + ar2 + ...... + arn-1 = \frac{a(r^n-1)}{r-1}
**Solution:
We have,
P(n) = a + ar + ar2 + ...... + arn-1 = \frac{a(r^n-1)}{r-1}
For n=1, we get
P(1) = a = \frac{a(r^1-1)}{r-1} = \frac{a(r-1)}{r-1} = a
So, P(1) is true
Assume that P(k) is true for some positive integer n=k
P(k) = a + ar + ar2 + ...... + ark-1 = \frac{a(r^k-1)}{r-1} .................(1)
Let's prove that P(k + 1) is also true. Now, we have
P(k+1) = a + ar + ar2 + ...... + ark-1 + ar(k+1)-1
= (a + ar + ar2 + ...... + ark-1) + ark
From Eq(1), we get
= \frac{a(r^k-1)}{r-1} + ark
= \frac{a(r^k-1) + ar^k(r-1)}{r-1}
= \frac{a(r^k-1 + r^{k+1}-r^k)}{r-1}
= \frac{a(r^{k+1}-1)}{r-1}
Hence,
P(k+1) = \frac{a(r^{k+1}-1)}{r-1}
**Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is **true for all natural numbers n.
Question 13: (1+ \frac{3}{1} ) (1+ \frac{5}{4} ) (1+ \frac{7}{9} ) ..... (1+ \frac{(2n+1)}{n^2} ) = (n+1)2
**Solution:
We have,
P(n) = (1+ \frac{3}{1}) (1+ \frac{5}{4}) (1+ \frac{7}{9}) ..... (1+\frac{(2n+1)}{n^2}) = (n+1)2
For **n=1, we get
P(1) = 1+ \frac{3}{1} = 1+3 = 4 = (1+1)2 = 22 = 4
So, P(1) is true
Assume that P(k) is true for some positive integer **n=k
P(k) = (1+ \frac{3}{1}) (1+ \frac{5}{4}) (1+ \frac{7}{9}) ..... (1+\frac{(2k+1)}{k^2}) = (k+1)2 .................(1)
Let's prove that P(k + 1) is also true. Now, we have
**P(k+1) = (1+ \frac{3}{1}) (1+ \frac{5}{4}) (1+ \frac{7}{9}) ..... (1+\frac{(2k+1)}{k^2}) (1+\frac{(2(k+1)+1)}{(k+1)^2})
= ((1+ \frac{3}{1}) (1+ \frac{5}{4}) (1+ \frac{7}{9}) ..... (1+\frac{(2k+1)}{k^2})) (1+\frac{2(k+1)+1}{(k+1)^2})
From Eq(1), we get
= (k+1)2 (1+\frac{2(k+1)+1}{(k+1)^2} )
= (k+1)2 (\frac{(k+1)^2 + 2(k+1)+1}{(k+1)^2})
= (k+1)2 + 2(k+1) + 1
= {(k+1)}2
Hence,
P(k+1) = {(k+1)}2
**Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is **true for all natural numbers n.