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 way. 100% reliable solutions for all the questions of SEBA Class 10 Advanced Maths Chapter 3.1
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Chapter 3.1 of SEBA Class 10 Advanced Mathematics
Chapter 3.2 of SEBA Class 10 Advanced Mathematics
Chapter 3.3 of SEBA Class 10 Advanced Mathematics
Chapter 3.4 of SEBA Class 10 Advanced Mathematics
Chapter 3.5 of SEBA Class 10 Advanced Mathematics
Q1: $1 + 3 + 5 + \dots + (2n – 1) = n^2$ 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 $n \in \mathbb{N}$.
Q2: 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 $n \in \mathbb{N}$.
Q3. 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 $n \in \mathbb{N}$.
Q4. 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 $n \in \mathbb{N}$.
Q5. 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 $n \in \mathbb{N}$.
Q6. 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 $n \in \mathbb{N}$.
Q7. 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 $n \in \mathbb{N}$.
Q8. 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 $S(k+1)$,
$= \underbrace{2 \cdot 6 \cdot 10 \cdot 14 \dots [4(k + 1) – 2]} \cdot (4k + 2)$
$= \underbrace{2 \cdot 6 \cdot 10 \cdot 14 \dots (4k – 2)} \cdot (4k + 2)$
$= \frac{\lfloor\underline{2k}}{\lfloor\underline{k}} \cdot (4k + 2)$, 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 $n \in \mathbb{N}$.
Q9. 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 $n \in \mathbb{N}$.
Q10. 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 $n \in \mathbb{N}$.
Q11. 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 $n \in \mathbb{N}$ ($n > 4$).
Q12. 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 $n \in \mathbb{N}$.