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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 1–10 (part 1 of 3)

  1. Exercise 1

    Give an example of a statement P(n)\displaystyle \mathrm{P}(n) which is true for all n≥4\displaystyle n \geq 4 but P(1),P(2)\displaystyle \mathrm{P}(1), \mathrm{P}(2) and P(3\displaystyle 3) are not true. Justify your answer.

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    NCERT’s answer
    \(\displaystyle \mathrm{P}(n): 2 n<\angle n\)
    Take \(\displaystyle P(n): 2n < n!\). It fails for \(\displaystyle n = 1, 2, 3\): \[2\cdot 1 = 2 > 1 = 1! \] \[2\cdot 2 = 4 > 2 = 2! \] \[2\cdot 3 = 6 = 3! \] It holds at \(\displaystyle n = 4\): \[2\cdot 4 = 8 < 24 = 4! \] Assume \(\displaystyle P(k)\) for some \(\displaystyle k \geq 4\): \[2k < k! \] Then \[2(k+1) = 2k + 2 < k! + 2 \] \[k! + 2 < k! + k\cdot k! \quad \text{(as } k\cdot k! \geq 96 > 2\text{)} \] \[k! + k\cdot k! = (k+1)\,k! = (k+1)! \] \[2(k+1) < (k+1)! \] So \(\displaystyle P(4)\) is true and \(\displaystyle P(k) \Rightarrow P(k+1)\) for \(\displaystyle k \geq 4\).Answer: \(\displaystyle P(n): 2n < n!\)

  2. Exercise 2

    Give an example of a statement P(n)\displaystyle \mathrm{P}(n) which is true for all n\displaystyle n. Justify your answer.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \mathrm{P}(n): 1+2+3+\ldots+n=\frac{n(n+1)}{2}\)
    Take \(\displaystyle P(n): 1 + 2 + 3 + \cdots + n = \dfrac{n(n+1)}{2}\). \[P(1): \; 1 = \frac{1\cdot 2}{2} \] Assume \(\displaystyle P(k)\): \[1 + 2 + \cdots + k = \frac{k(k+1)}{2} \] Add \(\displaystyle k+1\) to both sides: \[1 + 2 + \cdots + k + (k+1) = \frac{k(k+1)}{2} + (k+1) \] \[= \frac{(k+1)(k+2)}{2} \] So \(\displaystyle P(k+1)\) is true whenever \(\displaystyle P(k)\) is, and \(\displaystyle P(1)\) is true.Answer: \(\displaystyle P(n): 1 + 2 + 3 + \cdots + n = \dfrac{n(n+1)}{2}\)
  3. Prove each of the statements in Exercises $\displaystyle 3$ - $\displaystyle 16$ by the Principle of Mathematical Induction :

    Exercise 3

    4n−1\displaystyle 4^n-1 is divisible by 3\displaystyle 3, for each natural number n\displaystyle n.

    Not cross-checked

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

    Let \(\displaystyle P(n)\): \(\displaystyle 4^n - 1\) is divisible by 3. \[P(1): \; 4^1 - 1 = 3 \] Assume \(\displaystyle P(k)\): \[4^k - 1 = 3m, \quad m \in \mathbb{Z} \] Then \[4^{k+1} - 1 = 4\cdot 4^k - 1 = 4(3m+1) - 1 \] \[= 12m + 3 = 3(4m+1) \] So \(\displaystyle P(k+1)\) is true whenever \(\displaystyle P(k)\) is; by induction \(\displaystyle P(n)\) holds for all \(\displaystyle n \in \mathbb{N}\).Answer: \(\displaystyle 4^n - 1\) is divisible by $\displaystyle 3$ for every natural number \(\displaystyle n\).

  4. Exercise 4

    23n−1\displaystyle 2^{3 n}-1 is divisible by 7\displaystyle 7, for all natural numbers n\displaystyle n.

    Not cross-checked

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

    Let \(\displaystyle P(n)\): \(\displaystyle 2^{3n} - 1\) is divisible by 7. \[P(1): \; 2^3 - 1 = 7 \] Assume \(\displaystyle P(k)\): \[2^{3k} - 1 = 7m, \quad m \in \mathbb{Z} \] Then \[2^{3(k+1)} - 1 = 8\cdot 2^{3k} - 1 = 8(7m+1) - 1 \] \[= 56m + 7 = 7(8m+1) \] So \(\displaystyle P(k+1)\) is true whenever \(\displaystyle P(k)\) is; by induction \(\displaystyle P(n)\) holds for all \(\displaystyle n \in \mathbb{N}\).Answer: \(\displaystyle 2^{3n} - 1\) is divisible by $\displaystyle 7$ for every natural number \(\displaystyle n\).

  5. Exercise 5

    n3−7n+3\displaystyle n^3-7 n+3 is divisible by 3\displaystyle 3, for all natural numbers n\displaystyle n.

    Not cross-checked

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

    Let \(\displaystyle P(n)\): \(\displaystyle n^3 - 7n + 3\) is divisible by 3. \[P(1): \; 1 - 7 + 3 = -3 = 3(-1) \] Assume \(\displaystyle P(k)\): \[k^3 - 7k + 3 = 3m, \quad m \in \mathbb{Z} \] Then \[(k+1)^3 - 7(k+1) + 3 = k^3 + 3k^2 + 3k + 1 - 7k - 7 + 3 \] \[= (k^3 - 7k + 3) + 3k^2 + 3k - 6 \] \[= 3m + 3(k^2 + k - 2) = 3(m + k^2 + k - 2) \] So \(\displaystyle P(k+1)\) is true whenever \(\displaystyle P(k)\) is; by induction \(\displaystyle P(n)\) holds for all \(\displaystyle n \in \mathbb{N}\).Answer: \(\displaystyle n^3 - 7n + 3\) is divisible by $\displaystyle 3$ for every natural number \(\displaystyle n\).

  6. Exercise 6

    32n−1\displaystyle 3^{2 n}-1 is divisible by 8\displaystyle 8, for all natural numbers n\displaystyle n.

    Not cross-checked

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

    Let \(\displaystyle P(n)\): \(\displaystyle 3^{2n} - 1\) is divisible by 8. \[P(1): \; 3^2 - 1 = 8 \] Assume \(\displaystyle P(k)\): \[3^{2k} - 1 = 8m, \quad m \in \mathbb{Z} \] Then \[3^{2(k+1)} - 1 = 9\cdot 3^{2k} - 1 = 9(8m+1) - 1 \] \[= 72m + 8 = 8(9m+1) \] So \(\displaystyle P(k+1)\) is true whenever \(\displaystyle P(k)\) is; by induction \(\displaystyle P(n)\) holds for all \(\displaystyle n \in \mathbb{N}\).Answer: \(\displaystyle 3^{2n} - 1\) is divisible by $\displaystyle 8$ for every natural number \(\displaystyle n\).

  7. Exercise 7

    For any natural number n,7n−2n\displaystyle n, 7^n-2^n is divisible by 5.

    Not cross-checked

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

    Let \(\displaystyle P(n)\): \(\displaystyle 7^n - 2^n\) is divisible by 5. \[P(1): \; 7 - 2 = 5 \] Assume \(\displaystyle P(k)\): \[7^k - 2^k = 5m, \quad m \in \mathbb{Z} \] Then \[7^{k+1} - 2^{k+1} = 7\cdot 7^k - 2\cdot 2^k \] \[= 7(7^k - 2^k) + 7\cdot 2^k - 2\cdot 2^k \] \[= 7\cdot 5m + 5\cdot 2^k = 5(7m + 2^k) \] So \(\displaystyle P(k+1)\) is true whenever \(\displaystyle P(k)\) is; by induction \(\displaystyle P(n)\) holds for all \(\displaystyle n \in \mathbb{N}\).Answer: \(\displaystyle 7^n - 2^n\) is divisible by $\displaystyle 5$ for every natural number \(\displaystyle n\).

  8. Exercise 8

    For any natural number n,xn−yn\displaystyle n, x^n-y^n is divisible by x−y\displaystyle x-y, where x\displaystyle x and y\displaystyle y are any integers with x≠y\displaystyle x \neq y.

    Not cross-checked

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

    Let \(\displaystyle P(n)\): \(\displaystyle x^n - y^n\) is divisible by \(\displaystyle x - y\). \[P(1): \; x^1 - y^1 = x - y \] Assume \(\displaystyle P(k)\): \[x^k - y^k = (x-y)m, \quad m \in \mathbb{Z} \] Then \[x^{k+1} - y^{k+1} = x\cdot x^k - y\cdot y^k \] \[= x(x^k - y^k) + x y^k - y\cdot y^k \] \[= x(x-y)m + y^k(x-y) = (x-y)(xm + y^k) \] So \(\displaystyle P(k+1)\) is true whenever \(\displaystyle P(k)\) is; by induction \(\displaystyle P(n)\) holds for all \(\displaystyle n \in \mathbb{N}\).Answer: \(\displaystyle x^n - y^n\) is divisible by \(\displaystyle x - y\) for every natural number \(\displaystyle n\).

  9. Exercise 9

    n3−n\displaystyle n^3-n is divisible by 6\displaystyle 6, for each natural number n≥2\displaystyle n \geq 2.

    Not cross-checked

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

    Basis \(\displaystyle n=2\): \[2^3-2=6 \] Assume true for \(\displaystyle n=k\ge 2\): \[k^3-k=6m \] Step \(\displaystyle n=k+1\): \[(k+1)^3-(k+1)=k^3+3k^2+2k=(k^3-k)+3k(k+1) \] \[k(k+1)=2r \quad \text{(product of consecutive integers is even)} \] \[(k+1)^3-(k+1)=6m+6r=6(m+r) \] Answer: true for \(\displaystyle n=2\), and \(\displaystyle n=k\Rightarrow n=k+1\); so \(\displaystyle n^3-n\) is divisible by $\displaystyle 6$ for every \(\displaystyle n\ge 2\).

  10. Exercise 10

    n(n2+5)\displaystyle n\left(n^2+5\right) is divisible by 6\displaystyle 6, for each natural number n\displaystyle n.

    Not cross-checked

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

    Basis \(\displaystyle n=1\): \[1\cdot(1+5)=6 \] Assume true for \(\displaystyle n=k\): \[k(k^2+5)=6m \] Step \(\displaystyle n=k+1\): \[(k+1)\left[(k+1)^2+5\right]=k^3+3k^2+8k+6 \] \[=(k^3+5k)+3k(k+1)+6 \] \[k(k+1)=2r \quad \text{(product of consecutive integers is even)} \] \[(k+1)\left[(k+1)^2+5\right]=6m+6r+6=6(m+r+1) \] Answer: true for \(\displaystyle n=1\), and \(\displaystyle n=k\Rightarrow n=k+1\); so \(\displaystyle n(n^2+5)\) is divisible by $\displaystyle 6$ for every natural \(\displaystyle n\).