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NCERT Exemplar · Class 11 Mathematics Principle of Mathematical Induction

30 questions · 30 still being checked

This chapter is from the older syllabus and is not in the current NCERT textbook, so you may skip it.

EXERCISE 4.3 11–20 (part 2 of 3)

  1. Prove each of the statements in Exercises $\displaystyle 3$ - $\displaystyle 16$ by the Principle of Mathematical Induction :

    Exercise 11

    n2<2n\displaystyle n^2<2^n for all natural numbers n≥5\displaystyle n \geq 5.

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    Basis \(\displaystyle n=5\): \[5^2=25<32=2^5 \] Assume true for \(\displaystyle n=k\ge 5\): \[k^2<2^k \] Step \(\displaystyle n=k+1\): \[2^{k+1}=2\cdot 2^k>2k^2 \] \[2k^2-(k+1)^2=k^2-2k-1=(k-1)^2-2>0 \quad (k\ge 5) \] \[2^{k+1}>2k^2>(k+1)^2 \] Answer: true for \(\displaystyle n=5\), and \(\displaystyle n=k\Rightarrow n=k+1\); so \(\displaystyle n^2<2^n\) for every \(\displaystyle n\ge 5\).

  2. Exercise 12

    2n<(n+2)!\displaystyle 2 n<(n+2)! for all natural number n\displaystyle n.

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    Basis \(\displaystyle n=1\): \[2\cdot 1=2<6=3! \] Assume true for \(\displaystyle n=k\): \[2k<(k+2)! \] Step \(\displaystyle n=k+1\): \[(k+3)!=(k+3)\,(k+2)!>(k+3)\cdot 2k \] \[2k(k+3)-2(k+1)=2\left(k^2+2k-1\right)>0 \quad (k\ge 1) \] \[(k+3)!>2k(k+3)>2(k+1) \] Answer: true for \(\displaystyle n=1\), and \(\displaystyle n=k\Rightarrow n=k+1\); so \(\displaystyle 2n<(n+2)!\) for every natural \(\displaystyle n\).

  3. Exercise 13

    n<11+12+…+1n\displaystyle \sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}, for all natural numbers n≥2\displaystyle n \geq 2.

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    Basis \(\displaystyle n=2\): \[1+\frac{1}{\sqrt2}-\sqrt2=1-\frac{1}{\sqrt2}>0 \quad\Rightarrow\quad \sqrt2<1+\frac{1}{\sqrt2} \] Assume true for \(\displaystyle n=k\ge 2\): \[\sqrt{k}<\frac{1}{\sqrt1}+\frac{1}{\sqrt2}+\ldots+\frac{1}{\sqrt k} \] Step \(\displaystyle n=k+1\), add \(\displaystyle \frac{1}{\sqrt{k+1}}\) to both sides: \[\frac{1}{\sqrt1}+\ldots+\frac{1}{\sqrt{k+1}}>\sqrt{k}+\frac{1}{\sqrt{k+1}}=\frac{\sqrt{k(k+1)}+1}{\sqrt{k+1}} \] \[\sqrt{k(k+1)}>\sqrt{k^2}=k \] \[\frac{\sqrt{k(k+1)}+1}{\sqrt{k+1}}>\frac{k+1}{\sqrt{k+1}}=\sqrt{k+1} \] Answer: true for \(\displaystyle n=2\), and \(\displaystyle n=k\Rightarrow n=k+1\); so the inequality holds for every \(\displaystyle n\ge 2\).

  4. Exercise 14

    2+4+6+…+2n=n2+n\displaystyle 2+4+6+\ldots+2 n=n^2+n for all natural numbers n\displaystyle n.

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    Basis \(\displaystyle n=1\): \[2=1^2+1 \] Assume true for \(\displaystyle n=k\): \[2+4+6+\ldots+2k=k^2+k \] Step \(\displaystyle n=k+1\), add \(\displaystyle 2(k+1)\): \[2+4+\ldots+2k+2(k+1)=k^2+k+2k+2 \] \[=k^2+3k+2=(k+1)^2+(k+1) \] Answer: true for \(\displaystyle n=1\), and \(\displaystyle n=k\Rightarrow n=k+1\); so \(\displaystyle 2+4+\ldots+2n=n^2+n\) for every natural \(\displaystyle n\).

  5. Exercise 15

    1+2+22+…+2n=2n+1−1\displaystyle 1+2+2^2+\ldots+2^n=2^{n+1}-1 for all natural numbers n\displaystyle n.

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    Basis \(\displaystyle n=1\): \[1+2=3=2^2-1 \] Assume true for \(\displaystyle n=k\): \[1+2+2^2+\ldots+2^k=2^{k+1}-1 \] Step \(\displaystyle n=k+1\), add \(\displaystyle 2^{k+1}\): \[1+2+\ldots+2^k+2^{k+1}=2^{k+1}-1+2^{k+1} \] \[=2\cdot 2^{k+1}-1=2^{k+2}-1 \] Answer: true for \(\displaystyle n=1\), and \(\displaystyle n=k\Rightarrow n=k+1\); so \(\displaystyle 1+2+\ldots+2^n=2^{n+1}-1\) for every natural \(\displaystyle n\).

  6. Exercise 16

    1+5+9+…+(4n−3)=n(2n−1)\displaystyle 1+5+9+\ldots+(4 n-3)=n(2 n-1) for all natural numbers n\displaystyle n.

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    Basis \(\displaystyle n=1\): \[1=1\cdot(2\cdot1-1) \] Assume true for \(\displaystyle n=k\): \[1+5+9+\ldots+(4k-3)=k(2k-1) \] Step \(\displaystyle n=k+1\), add \(\displaystyle 4(k+1)-3=4k+1\): \[1+5+\ldots+(4k-3)+(4k+1)=2k^2-k+4k+1 \] \[=2k^2+3k+1=(k+1)(2k+1)=(k+1)\left[2(k+1)-1\right] \] Answer: true for \(\displaystyle n=1\), and \(\displaystyle n=k\Rightarrow n=k+1\); so \(\displaystyle 1+5+9+\ldots+(4n-3)=n(2n-1)\) for every natural \(\displaystyle n\).
  7. Use the Principle of Mathematical Induction in the following Exercises.

    Exercise 17

    A sequence a1,a2,a3…\displaystyle a_1, a_2, a_3 \ldots is defined by letting a1=3\displaystyle a_1=3 and ak=7ak−1\displaystyle a_k=7 a_{k-1} for all natural numbers k≥2\displaystyle k \geq 2. Show that an=3.7n−1\displaystyle a_n=3.7^{n-1} for all natural numbers.

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    Let \(\displaystyle P(n)\): \(\displaystyle a_n = 3\cdot 7^{n-1}\). Base case: \[a_1 = 3 = 3\cdot 7^{0} \] Step: assume \(\displaystyle P(k)\), i.e. \(\displaystyle a_k = 3\cdot 7^{k-1}\). \[a_{k+1} = 7a_k \quad \text{(definition)} \] \[= 7\cdot 3\cdot 7^{k-1} \quad \text{(hypothesis)} \] \[= 3\cdot 7^{k} = 3\cdot 7^{(k+1)-1} \] Answer: \(\displaystyle P(1)\) is true and \(\displaystyle P(k)\Rightarrow P(k+1)\), so \(\displaystyle a_n = 3\cdot 7^{n-1}\) for all \(\displaystyle n\in\mathbf N\).

  8. Exercise 18

    A sequence b0,b1,b2…\displaystyle b_0, b_1, b_2 \ldots is defined by letting b0=5\displaystyle b_0=5 and bk=4+bk−1\displaystyle b_k=4+b_{k-1} for all natural numbers k\displaystyle k. Show that bn=5+4n\displaystyle b_n=5+4 n for all natural number n\displaystyle n using mathematical induction.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Let \(\displaystyle P(n)\): \(\displaystyle b_n = 5+4n\). Base case: \[b_0 = 5 = 5+4\cdot 0 \quad \text{(given)} \] \[b_1 = 4 + b_0 = 9 = 5+4\cdot 1 \] Step: assume \(\displaystyle P(k)\), i.e. \(\displaystyle b_k = 5+4k\). \[b_{k+1} = 4 + b_k \quad \text{(definition)} \] \[= 4 + 5 + 4k \quad \text{(hypothesis)} \] \[= 5 + 4(k+1) \] Answer: \(\displaystyle P\) holds at the start and passes from \(\displaystyle k\) to \(\displaystyle k+1\), so \(\displaystyle b_n = 5+4n\) for all \(\displaystyle n\).

  9. Exercise 19

    A sequence d1,d2,d3…\displaystyle d_1, d_2, d_3 \ldots is defined by letting d1=2\displaystyle d_1=2 and dk=dk−1k\displaystyle d_k=\frac{d_{k-1}}{k} for all natural numbers, k≥2\displaystyle k \geq 2. Show that dn=2n!\displaystyle d_n=\frac{2}{n!} for all n∈N\displaystyle n \in \mathbf{N}.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Let \(\displaystyle P(n)\): \(\displaystyle d_n = \dfrac{2}{n!}\). Base case: \[d_1 = 2 = \frac{2}{1!} \] Step: assume \(\displaystyle P(k)\), i.e. \(\displaystyle d_k = \dfrac{2}{k!}\). \[d_{k+1} = \frac{d_k}{k+1} \quad \text{(definition)} \] \[= \frac{2}{k!\,(k+1)} \quad \text{(hypothesis)} \] \[= \frac{2}{(k+1)!} \] Answer: \(\displaystyle P(1)\) is true and \(\displaystyle P(k)\Rightarrow P(k+1)\), so \(\displaystyle d_n = \dfrac{2}{n!}\) for all \(\displaystyle n\in\mathbf N\).

  10. Exercise 20

    Prove that for all n∈N\displaystyle n \in \mathbf{N} cos⁡α+cos⁡(α+β)+cos⁡(α+2β)+…+cos⁡(α+(n−1)β)=cos⁡(α+(n−12)β)sin⁡(nβ2)sin⁡β2\begin{gathered} \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(n-1) \beta) \\ =\frac{\cos \left(\alpha+\left(\dfrac{n-1}{2}\right) \beta\right) \sin \left(\dfrac{n \beta}{2}\right)}{\sin \dfrac{\beta}{2}} \end{gathered}

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    Let \(\displaystyle S_n\) be the left side and multiply the identity by \(\displaystyle \sin\frac{\beta}{2}\). Let \(\displaystyle P(n)\): \[S_n\sin\tfrac{\beta}{2} = \cos\!\Big(\alpha+\tfrac{(n-1)\beta}{2}\Big)\sin\tfrac{n\beta}{2} \] Base case: \[S_1\sin\tfrac{\beta}{2} = \cos\alpha\,\sin\tfrac{\beta}{2} = \cos\!\Big(\alpha+0\Big)\sin\tfrac{1\cdot\beta}{2} \] Step: assume \(\displaystyle P(k)\). Since \(\displaystyle S_{k+1} = S_k + \cos(\alpha+k\beta)\): \[S_{k+1}\sin\tfrac{\beta}{2} = \cos\!\Big(\alpha+\tfrac{(k-1)\beta}{2}\Big)\sin\tfrac{k\beta}{2} + \cos(\alpha+k\beta)\sin\tfrac{\beta}{2} \quad \text{(hypothesis)} \] Use \(\displaystyle 2\cos A\sin B = \sin(A+B)-\sin(A-B)\) on each term: \[= \tfrac12\Big[\sin\!\Big(\alpha+\tfrac{(2k-1)\beta}{2}\Big)-\sin\!\Big(\alpha-\tfrac{\beta}{2}\Big)\Big] + \tfrac12\Big[\sin\!\Big(\alpha+\tfrac{(2k+1)\beta}{2}\Big)-\sin\!\Big(\alpha+\tfrac{(2k-1)\beta}{2}\Big)\Big] \] \[= \tfrac12\Big[\sin\!\Big(\alpha+\tfrac{(2k+1)\beta}{2}\Big)-\sin\!\Big(\alpha-\tfrac{\beta}{2}\Big)\Big] \] \[= \cos\!\Big(\alpha+\tfrac{k\beta}{2}\Big)\sin\tfrac{(k+1)\beta}{2} \quad \text{(same identity, } A=\alpha+\tfrac{k\beta}{2},\ B=\tfrac{(k+1)\beta}{2}\text{)} \] Answer: \(\displaystyle P(1)\) is true and \(\displaystyle P(k)\Rightarrow P(k+1)\); dividing by \(\displaystyle \sin\frac{\beta}{2}\) gives the stated identity for all \(\displaystyle n\in\mathbf N\).