Class 10 Advanced Maths Chapter 3.1 Solutions: Arithmetic of Integers | For SEBA
In this article, we have solved all the questions from Chapter 3.1 of SEBA Class 10 Advanced Mathematics in a simple, uniform step-by-step manner.
Question1: $1 + 3 + 5 + \dots + (2n – 1) = n^2$ for all $n \in \mathbb{N}$.
Solution:
Let $S(n): 1 + 3 + 5 + \dots + (2n – 1) = n^2$.
$S(1): 1 = 1^2 = 1$, which is true.
Let $S(k): 1 + 3 + 5 + \dots + (2k – 1) = k^2$, is true for some $k \in \mathbb{N}$.
We need to show that,
$S(k+1): 1 + 3 + 5 + \dots + (2k – 1) + [2(k + 1) – 1] = (k + 1)^2$, is also true.
Now, LHS of $S(k+1)$,
$= \underbrace{1 + 3 + 5 + \dots + (2k – 1)} + [2(k + 1) – 1]$
$= k^2 + (2k + 2 – 1)$, Since $S(k)$ is true
$= k^2 + (2k + 1)$
$= k^2 + 2k + 1$
$= (k + 1)^2 = \text{RHS of } S(k+1)$.
Therefore, $S(k+1)$ is true if $S(k)$ is true.
By the method of mathematical induction, $S(n)$ is true for all $n \in \mathbb{N}$.
Question 2:
Show that: $2 + 4 + 6 + \dots + 2n = n(n + 1)$ for all $n \in \mathbb{N}$.
Solution:
Let $S(n): 2 + 4 + 6 + \dots + 2n = n(n + 1)$.
$S(1): 2 = 1(1 + 1) = 2$, which is true.
Let $S(k): 2 + 4 + 6 + \dots + 2k = k(k + 1)$, is true for some $k \in \mathbb{N}$.
We need to show that,
$S(k+1): 2 + 4 + 6 + \dots + 2k + 2(k + 1) = (k + 1)[(k + 1) + 1]$, is also true.
Now, LHS of $S(k+1)$,
$= \underbrace{2 + 4 + 6 + \dots + 2k} + 2(k + 1)$
$= k(k + 1) + 2(k + 1)$, Since $S(k)$ is true
$= (k + 1)(k + 2)$
$= (k + 1)[(k + 1) + 1] = \text{RHS of } S(k+1)$.
Therefore, $S(k+1)$ is true if $S(k)$ is true.
By the method of mathematical induction, $S(n)$ is true for all $n \in \mathbb{N}$.
Question 3
Show that: $1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$ for all $n \in \mathbb{N}$.
Solution:
Let $S(n): 1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$.
$S(1): 1^2 = \frac{1(1 + 1)(2 \cdot 1 + 1)}{6} = \frac{1 \cdot 2 \cdot 3}{6} = 1$, which is true.
Let $S(k): 1^2 + 2^2 + 3^2 + \dots + k^2 = \frac{k(k + 1)(2k + 1)}{6}$, is true for some $k \in \mathbb{N}$.
We need to show that,
$S(k+1): 1^2 + 2^2 + 3^2 + \dots + k^2 + (k + 1)^2 = \frac{(k + 1)[(k + 1) + 1][2(k + 1) + 1]}{6}$, is also true.
Now, LHS of $S(k+1)$,
$= \underbrace{1^2 + 2^2 + 3^2 + \dots + k^2} + (k + 1)^2$
$= \frac{k(k + 1)(2k + 1)}{6} + (k + 1)^2$, Since $S(k)$ is true
$= (k + 1) \left[ \frac{k(2k + 1)}{6} + (k + 1) \right]$
$= (k + 1) \left[ \frac{2k^2 + k + 6k + 6}{6} \right]$
$= (k + 1) \left[ \frac{2k^2 + 7k + 6}{6} \right]$
$= (k + 1) \left[ \frac{2k^2 + 4k + 3k + 6}{6} \right]$
$= (k + 1) \left[ \frac{2k(k + 2) + 3(k + 2)}{6} \right]$
$= \frac{(k + 1)(k + 2)(2k + 3)}{6}$
$= \frac{(k + 1)[(k + 1) + 1][2(k + 1) + 1]}{6} = \text{RHS of } S(k+1)$.
Therefore, $S(k+1)$ is true if $S(k)$ is true.
By the method of mathematical induction, $S(n)$ is true for all $n \in \mathbb{N}$.
Question 4
Show that: $1^2 + 3^2 + 5^2 + \dots + (2n – 1)^2 = \frac{n(2n – 1)(2n + 1)}{3}$ for all $n \in \mathbb{N}$.
Solution:
Let $S(n): 1^2 + 3^2 + 5^2 + \dots + (2n – 1)^2 = \frac{n(2n – 1)(2n + 1)}{3}$.
$S(1): 1^2 = \frac{1(2 \cdot 1 – 1)(2 \cdot 1 + 1)}{3} = \frac{1 \cdot 1 \cdot 3}{3} = 1$, which is true.
Let $S(k): 1^2 + 3^2 + 5^2 + \dots + (2k – 1)^2 = \frac{k(2k – 1)(2k + 1)}{3}$, is true for some $k \in \mathbb{N}$.
We need to show that,
$S(k+1): 1^2 + 3^2 + 5^2 + \dots + (2k – 1)^2 + [2(k + 1) – 1]^2 = \frac{(k + 1)[2(k + 1) – 1][2(k + 1) + 1]}{3}$, is also true.
Now, LHS of $S(k+1)$,
$= \underbrace{1^2 + 3^2 + 5^2 + \dots + (2k – 1)^2} + (2k + 1)^2$
$= \frac{k(2k – 1)(2k + 1)}{3} + (2k + 1)^2$, Since $S(k)$ is true
$= (2k + 1) \left[ \frac{k(2k – 1)}{3} + (2k + 1) \right]$
$= (2k + 1) \left[ \frac{2k^2 – k + 6k + 3}{3} \right]$
$= (2k + 1) \left[ \frac{2k^2 + 5k + 3}{3} \right]$
$= (2k + 1) \left[ \frac{2k^2 + 2k + 3k + 3}{3} \right]$
$= (2k + 1) \left[ \frac{2k(k + 1) + 3(k + 1)}{3} \right]$
$= \frac{(2k + 1)(k + 1)(2k + 3)}{3}$
$= \frac{(k + 1)(2k + 1)(2k + 3)}{3} = \text{RHS of } S(k+1)$.
Therefore, $S(k+1)$ is true if $S(k)$ is true.
By the method of mathematical induction, $S(n)$ is true for all $n \in \mathbb{N}$.
Question 5
Show that: $1^3 + 2^3 + 3^3 + \dots + n^3 = \left\{ \frac{n(n + 1)}{2} \right\}^2$ for all $n \in \mathbb{N}$.
Solution:
Let $S(n): 1^3 + 2^3 + 3^3 + \dots + n^3 = \left\{ \frac{n(n + 1)}{2} \right\}^2$.
$S(1): 1^3 = \left\{ \frac{1(1 + 1)}{2} \right\}^2 = 1^2 = 1$, which is true.
Let $S(k): 1^3 + 2^3 + 3^3 + \dots + k^3 = \left\{ \frac{k(k + 1)}{2} \right\}^2$, is true for some $k \in \mathbb{N}$.
We need to show that,
$S(k+1): 1^3 + 2^3 + 3^3 + \dots + k^3 + (k + 1)^3 = \left\{ \frac{(k + 1)[(k + 1) + 1]}{2} \right\}^2$, is also true.
Now, LHS of $S(k+1)$,
$= \underbrace{1^3 + 2^3 + 3^3 + \dots + k^3} + (k + 1)^3$
$= \frac{k^2(k + 1)^2}{4} + (k + 1)^3$, Since $S(k)$ is true
$= (k + 1)^2 \left[ \frac{k^2}{4} + (k + 1) \right]$
$= (k + 1)^2 \left[ \frac{k^2 + 4k + 4}{4} \right]$
$= (k + 1)^2 \frac{(k + 2)^2}{4}$
$= \left\{ \frac{(k + 1)(k + 2)}{2} \right\}^2$
$= \left\{ \frac{(k + 1)[(k + 1) + 1]}{2} \right\}^2 = \text{RHS of } S(k+1)$.
Therefore, $S(k+1)$ is true if $S(k)$ is true.
By the method of mathematical induction, $S(n)$ is true for all $n \in \mathbb{N}$.
Question 6
Show that: $2 + 2^2 + 2^3 + \dots + 2^n = 2(2^n – 1)$ for all $n \in \mathbb{N}$.
Solution:
Let $S(n): 2 + 2^2 + 2^3 + \dots + 2^n = 2(2^n – 1)$.
$S(1): 2 = 2(2^1 – 1) = 2(1) = 2$, which is true.
Let $S(k): 2 + 2^2 + 2^3 + \dots + 2^k = 2(2^k – 1)$, is true for some $k \in \mathbb{N}$.
We need to show that,
$S(k+1): 2 + 2^2 + 2^3 + \dots + 2^k + 2^{k+1} = 2(2^{k+1} – 1)$, is also true.
Now, LHS of $S(k+1)$,
$= \underbrace{2 + 2^2 + 2^3 + \dots + 2^k} + 2^{k+1}$
$= 2(2^k – 1) + 2^{k+1}$, Since $S(k)$ is true
$= 2 \cdot 2^k – 2 + 2^{k+1}$
$= 2^{k+1} + 2^{k+1} – 2$
$= 2 \cdot 2^{k+1} – 2$
$= 2(2^{k+1} – 1) = \text{RHS of } S(k+1)$.
Therefore, $S(k+1)$ is true if $S(k)$ is true.
By the method of mathematical induction, $S(n)$ is true for all $n \in \mathbb{N}$.
Question 7
Show that: $1\lfloor\underline{1} + 2\lfloor\underline{2} + 3\lfloor\underline{3} + \dots + n\lfloor\underline{n} = \lfloor\underline{n+1} – 1$ for all $n \in \mathbb{N}$.
Solution:
Let $S(n): 1\lfloor\underline{1} + 2\lfloor\underline{2} + 3\lfloor\underline{3} + \dots + n\lfloor\underline{n} = \lfloor\underline{n+1} – 1$.
$S(1): 1\lfloor\underline{1} = \lfloor\underline{1+1} – 1 = 2 – 1 = 1$, which is true.
Let $S(k): 1\lfloor\underline{1} + 2\lfloor\underline{2} + \dots + k\lfloor\underline{k} = \lfloor\underline{k+1} – 1$, is true for some $k \in \mathbb{N}$.
We need to show that,
$S(k+1): 1\lfloor\underline{1} + 2\lfloor\underline{2} + \dots + k\lfloor\underline{k} + (k + 1)\lfloor\underline{k+1} = \lfloor\underline{(k+1)+1} – 1$, is also true.
Now, LHS of $S(k+1)$,
$= \underbrace{1\lfloor\underline{1} + 2\lfloor\underline{2} + \dots + k\lfloor\underline{k}} + (k + 1)\lfloor\underline{k+1}$
$= \lfloor\underline{k+1} – 1 + (k + 1)\lfloor\underline{k+1}$, Since $S(k)$ is true
$= \lfloor\underline{k+1}[1 + (k + 1)] – 1$
$= \lfloor\underline{k+1}(k + 2) – 1$
$= \lfloor\underline{k+2} – 1$
$= \lfloor\underline{(k+1)+1} – 1 = \text{RHS of } S(k+1)$.
Therefore, $S(k+1)$ is true if $S(k)$ is true.
By the method of mathematical induction, $S(n)$ is true for all $n \in \mathbb{N}$.
Question 8
Show that: $2 \cdot 6 \cdot 10 \cdot 14 \dots (4n – 2) = \frac{\lfloor\underline{2n}}{\lfloor\underline{n}}$ for all $n \in \mathbb{N}$.
Solution:
Let $S(n): 2 \cdot 6 \cdot 10 \cdot 14 \dots (4n – 2) = \frac{\lfloor\underline{2n}}{\lfloor\underline{n}}$.
$S(1): 2 = \frac{\lfloor\underline{2 \cdot 1}}{\lfloor\underline{1}} = \frac{2}{1} = 2$, which is true.
Let $S(k): 2 \cdot 6 \cdot 10 \cdot 14 \dots (4k – 2) = \frac{\lfloor\underline{2k}}{\lfloor\underline{k}}$, is true for some $k \in \mathbb{N}$.
We need to show that,
$S(k+1): 2 \cdot 6 \cdot 10 \cdot 14 \dots (4k – 2) \cdot [4(k + 1) – 2] = \frac{\lfloor\underline{2(k+1)}}{\lfloor\underline{k+1}}$, is also true.
Now, LHS of $S(k+1)$,
$= \underbrace{2 \cdot 6 \cdot 10 \cdot 14 \dots (4k – 2)} \cdot (4k + 2)$
$= \frac{\lfloor\underline{2k}}{\lfloor\underline{k}} \cdot (4k + 2)$, Since $S(k)$ is true
$= \frac{\lfloor\underline{2k}}{\lfloor\underline{k}} \cdot 2(2k + 1)$
$= \frac{\lfloor\underline{2k} \cdot 2(2k + 1) \cdot (k + 1)}{\lfloor\underline{k} \cdot (k + 1)}$ (Multiplying numerator & denominator by $k+1$)
$= \frac{\lfloor\underline{2k} \cdot (2k + 1) \cdot (2k + 2)}{\lfloor\underline{k+1}}$
$= \frac{\lfloor\underline{2k+2}}{\lfloor\underline{k+1}}$
$= \frac{\lfloor\underline{2(k+1)}}{\lfloor\underline{k+1}} = \text{RHS of } S(k+1)$.
Therefore, $S(k+1)$ is true if $S(k)$ is true.
By the method of mathematical induction, $S(n)$ is true for all $n \in \mathbb{N}$.
Question 9
Show that: $(2n + 7) < (n + 3)^2$ for all $n \in \mathbb{N}$.
Solution:
Let $S(n): (2n + 7) < (n + 3)^2$.
$S(1): (2 \cdot 1 + 7) < (1 + 3)^2 \implies 9 < 16$, which is true.
Let $S(k): (2k + 7) < (k + 3)^2$, is true for some $k \in \mathbb{N}$.
We need to show that,
$S(k+1): [2(k + 1) + 7] < [(k + 1) + 3]^2 \implies (2k + 9) < (k + 4)^2$, is also true.
Now, LHS of $S(k+1)$,
$= 2k + 9$
$= (2k + 7) + 2$
$< (k + 3)^2 + 2$, Since $S(k)$ is true
$= k^2 + 6k + 9 + 2$
$= k^2 + 6k + 11$
$< k^2 + 8k + 16$, Since $k \in \mathbb{N} \implies 6k < 8k$ and $11 < 16$
$= (k + 4)^2$
$= [(k + 1) + 3]^2 = \text{RHS of } S(k+1)$.
Therefore, $S(k+1)$ is true if $S(k)$ is true.
By the method of mathematical induction, $S(n)$ is true for all $n \in \mathbb{N}$.
Question 10
Show that: $2^n > n$ for all $n \in \mathbb{N}$.
Solution:
Let $S(n): 2^n > n$.
$S(1): 2^1 > 1 \implies 2 > 1$, which is true.
Let $S(k): 2^k > k$, is true for some $k \in \mathbb{N}$.
We need to show that,
$S(k+1): 2^{k+1} > k + 1$, is also true.
Now, LHS of $S(k+1)$,
$= 2^{k+1}$
$= 2 \cdot 2^k$
$> 2k$, Since $S(k)$ is true
$= k + k$
$\ge k + 1$, Since $k \in \mathbb{N} \implies k \ge 1$
$= \text{RHS of } S(k+1)$.
Therefore, $S(k+1)$ is true if $S(k)$ is true.
By the method of mathematical induction, $S(n)$ is true for all $n \in \mathbb{N}$.
Question 11
Show that: $\lfloor\underline{n} > 2^n$ for all $n \in \mathbb{N}$ and $n > 4$.
Solution:
Let $S(n): \lfloor\underline{n} > 2^n$.
Since $n > 4$, the initial case is $n = 5$.
$S(5): \lfloor\underline{5} > 2^5 \implies 120 > 32$, which is true.
Let $S(k): \lfloor\underline{k} > 2^k$, is true for some $k \in \mathbb{N}$ where $k \ge 5$.
We need to show that,
$S(k+1): \lfloor\underline{k+1} > 2^{k+1}$, is also true.
Now, LHS of $S(k+1)$,
$= \lfloor\underline{k+1}$
$= (k + 1)\lfloor\underline{k}$
$> (k + 1)2^k$, Since $S(k)$ is true
$> 2 \cdot 2^k$, Since $k \ge 5 \implies k + 1 > 2$
$= 2^{k+1} = \text{RHS of } S(k+1)$.
Therefore, $S(k+1)$ is true if $S(k)$ is true.
By the method of mathematical induction, $S(n)$ is true for all $n \in \mathbb{N}$ ($n > 4$).
Question 12
Show that: $(1 + a)^n \ge 1 + na$ for $1 + a > 0$ and $n \in \mathbb{N}$.
Solution:
Let $S(n): (1 + a)^n \ge 1 + na$.
$S(1): (1 + a)^1 \ge 1 + 1 \cdot a \implies 1 + a \ge 1 + a$, which is true.
Let $S(k): (1 + a)^k \ge 1 + ka$, is true for some $k \in \mathbb{N}$.
We need to show that,
$S(k+1): (1 + a)^{k+1} \ge 1 + (k + 1)a$, is also true.
Now, LHS of $S(k+1)$,
$= (1 + a)^{k+1}$
$= (1 + a)^k(1 + a)$
$\ge (1 + ka)(1 + a)$, Since $S(k)$ is true and $(1 + a) > 0$
$= 1 + a + ka + ka^2$
$= 1 + (k + 1)a + ka^2$
$\ge 1 + (k + 1)a$, Since $k \in \mathbb{N}$ and $a^2 \ge 0 \implies ka^2 \ge 0$
$= \text{RHS of } S(k+1)$.
Therefore, $S(k+1)$ is true if $S(k)$ is true.
By the method of mathematical induction, $S(n)$ is true for all $n \in \mathbb{N}$.