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Class 11 NCERT Solutions- Chapter 4 Principle of Mathematical Induction – Exercise 4.1 | Set 1

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  • Last Updated : 29 Nov, 2022
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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.


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