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- Question 12 Exercise 7.1
- n}-1}{24}$ is an integer. Solution: 1. For $n=1$, then $$\dfrac{5^{2 n}-1}{24}=\dfrac{5^{2.1}-1}{24}=\df... is true for $n=1$ 2. Let it be true for $n=k>1$ then $$\dfrac{5^{2 k}-1}{24} \in \mathbb{Z}$$ 3. For $n=k+1$ then consider \begin{align}\dfrac{5^{2(k+1)}-1}{24}&=\... 81}$ is an integer. ====Solution==== 1. For $n=1$ then \begin{align}\dfrac{10^{n+1}-9 n-10}{81}&=\dfrac{
- Question 13 Exercise 7.1
- n \in \mathbf{N}$. ====Solution==== 1. For $n=1$ then $2^n=2^1=2$ and $n=1$. Clearly $2>1$. hence the ... is true for $n=1$. 2. Let it be true for $n=l>I$ then $2^k>k\cdots(i)$ 3. For $n=k+1$ then we consider \begin{align} & 2^{k+1}=2^k \cdot 2>k \cdot 2 \quad \... integer $n \geq 4$ ====Solution==== 1. For $n=4$ then $n !=4 !=24$ and $n^2=4^2=16$. Clearly $24>16$,
- Question 14 Exercise 7.1
- y positive integer. ====Solution==== 1. For $n=1$ then $$3^{2 n-1}+2^{2 n-1}=3^{2.1-1}+2^{2.1-1}=5 \text... true for $n=1$ 2. Let it be true for $n=k>1$, then $54$ divides $3^{2 k} 1+2^{2 k} \quad 1$ which im... 1}=5 Q$$ where $Q$ is a quotient. 3. For $n=k+1$ then considering \begin{align} 3^{2(k+1)-1}+2^{2(k+1)-... ll natural numbers. ====Solution==== 1. For $n=1$ then $$2^{2 n}-1=2^{2.1}-1=4-1=3 $$ $3$ divides $3$, h
- Question 7 & 8 Review Exercise 7
- tion to prove the given statement. (1.) For $n=1$ then $7^k-3^k=7-4=4$. Thus 4 divides 4. Hence given is true for $n=1$. (2.) Let it be true for $n=k>1$ then $7^n-3^n=4 Q$ where $Q$ is the quotient in the induction hypothesis. (3.) For $n=k+1$ then we have $$ \begin{aligned} & 7^{k+1}-3^{k+1}=7.7^... this using mathernatical induction. 1. For $n=1$ then $$ (1+x)^1=1+x=1+1 x $$ Thus it is true for $n=1
- Question 7 Exercise 7.1
- ac{n(n+1)(n+2)}{3}$ ====Solution==== 1. For $n=1$ then $$1.2=2=\dfrac{1(1+1)(1+2)}{3}=2 $$ Thus it is true for $n=1$. 2. Let it be true for $n=k$, then \begin{align}1.2+2.3+3.4+\ldots+k(k+1)& =\dfrac{k... {3}....(i)\end{align} 3. Considering for $n=k+1$, then $(k-1)^{t h}$ term of the series on left is $a_{k
- Question 10 Exercise 7.1
- \end{array}\right)$ ====Solution==== 1. For $n=1$ then \begin{align}\left(\begin{array}{l}5 \\ 5\end{arr... it is true for $n=1$. 2. Let it be true for $n=k$ then \begin{align}\left(\begin{array}{l}5 \\ 5\end{arr... {array}\right) . ...(i)\end{align} 3. For $n=k+1$ then $(k+1)^{t h}$ term of the series on the left is $
- Question 11 Exercise 7.1
- \geq 2 \end{align} ====Solution==== 1. For $n=2$ then \begin{align} \left(\begin{array}{l} 2 \\ 2 \end{... t is true for $n=1$. 2. Let it be true for $n=k$ then $$\left(\begin{array}{l} 2 \\ 2 \end{array}\right... }{c} k-1 \\ 3 \end{array}\right)$$ 3. For $n=k+1$ then $(k+i)^{u / i}$ term of the series on the left is
- Question 15 Exercise 7.1
- =2 n, \quad m \in \mathbb{Z}^{+}$. 1. For $m=1$, then $$a^{2 n}-b^{2 m}=a^2-b^2=(a+b)(a-b)$$ $\Rightarr... for $m=1$ or $n=2$. 2. Let it be true for $m=k$ then $$a^{2 k}-b^{2 k}=Q(a+b)$$ where $Q$ is quotient in the induction hypothesis. 3. For $m=k+1$ then it becomes \begin{align}a^{2(k+1)}-b^{2(k-1)} & =
- Question 14 Exercise 7.3
- , Pakistan. Q14 If $x$ is nearly equal to unity, then show that $p x^p-q x^q=(p-q) x^{p+q}$. Solution: ... so let say $x=1+h$. where $h \longrightarrow 0$. Then $$ p x^p-q x^q=p(1+h)^p-q(1+h)^q $$ Applying bin... ^{p+q} . \end{aligned} $$ Replacing $1+h$ by $x$ then we have ====Go To==== <text align="left"><btn
- Question 1 Exercise 7.1
- +\cdots+2 n=n(n+1)$ ====Solution==== 1. For $n=1$ then $$2=1(1+1)=2 $$ Hence the above proposition is true for $n=1$. 2. Let it be true for $n=k$, then $$2+4+6+\cdots+2 k=k(k+1)....(i)$$ 3. For $n=k+1$
- Question 2 Exercise 7.1
- s+(4 n-3)=n(2 n-1)$ ====Solution==== 1. For $n=1$ then $$1=1(2.1-1)=1$$ Hence the above proposition is true for $n=1$. 2. Let it be true for $n=k$, then \begin{align}1+5+9+\ldots+(4 k-3)\\ & =k(2 k-1)..
- Question 5 Exercise 7.1
- +1)}{2}\right]^2$. ====Solution==== 1. For $n=1$, then $1^3=1=\left[\dfrac{1(1+1)}{2}\right]^2=1$. Thus... s true for $n=1$. 2. Let it be true for $n=k_1$, then \begin{align}1^3+2^3+3^3+\ldots+k^3& =[\dfrac{k(k
- Question 6 Exercise 7.1
- n !)= -(n+1) !-1$ ====Solution==== 1. For $n=1$, then $$1(1 !)=1=(1+1) !-1=2 !-1=1 $$ Thus it is truc for $n=1$. 2. Let it be true for $n=k$, then we have \begin{align}1(1 !)+2(2 !)+3(3 !)+\ldots+
- Question 8 Exercise 7.1
- is true for $n=1$. 2. Let it be true for $n-k>1$ then \begin{align}1+2+2^2+2^3+\ldots+2^{k-1} \\ & =2^k... -1 ....(i)\end{align} 3. Considering for $n-k-1$, then $(k+1)^{t h}$ term of the series on the left is $
- Question 9 Exercise 7.1
- [1-\dfrac{1}{3^n}]$ ====Solution==== 1. For $n=1$ then $$\dfrac{1}{3}-\dfrac{1}{2}[1-\dfrac{1}{3}]-\dfra... t is true for $n=1$. 2. Let it be true for $n=k$ then $$\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ldots+