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- Question 10 Exercise 7.3
- s \end{aligned} $$ Comparing both the series, we have $n x=-\frac{1}{4}$ (I) and $\frac{n(n-1)}{2 !} x^... \cdot \frac{1}{2^4}$ Taking square of Eq.(1), we have $n^2 x^2=\frac{1}{16}$ Dividing Eq.(2) by Eq.(3),... ts \end{aligned} $$ Comparing both the series, we have $$ \begin{aligned} & n x=\frac{5}{8} \\ & \frac{n... } . \end{aligned} $$ Taking square of Eq.(1), we have $$ n^2 x^2=\frac{25}{64} $$ Dividing Eq.(2) by E
- Question 10 Exercise 7.2
- \cdot $$ Putting $x=1$ in the above equation, we have $(1 \div 1)^n=\left(\begin{array}{l}n \\ 0\end{ar... (\cdots 1)^n\right. \\ & \end{aligned} $$ Vow we have two cases Case- 1 If $n$ is caen then $$ \begin{a... of even and odd chefficients are cyual. Nins we have shown that uncomplete question ./.;;./;;....
- Question 12 Exercise 7.3
- s \end{aligned} $$ Comparing both the series, we have $n x=\frac{1}{2^2}=\frac{1}{4}.... (1)$ and $$ \f... dot \frac{1}{2^4} $$ Taking square of Eq.(1), we have $n^2 x^2=\frac{1}{16}$ Dividing Eq.(2) by Eq.(3),... dot \frac{1}{2^4} $$ Taking square of Eq.(1), we have $n^2 x^2=\frac{1}{16}$ Dividing Eq.(2) by Eq.(3),
- Question 3 Exercise 7.1
- true for $n=1$. 2. Let it be true for $n=k$, we have $$3+6+9+\ldots+3 k=\dfrac{3 k(k+1)}{2}....(i)$$ 3... to both sides of the induction hypothesis (i), we have \begin{align}3+6+9+\ldots+3 k+3(k+1) & =\dfrac{3
- Question 4 Exercise 7.1
- true for $n=1$. 2. Let it be true for $n=k$, we have \begin{align}3+7+11+\cdots+(4 k-1) & =k(2 k+1)..... to both sides of the induction hypothesis (i), we have \begin{align} 3+7+11+\cdots+(4 k-1)+[4(k+1)-1] &
- Question 6 Exercise 7.1
- for $n=1$. 2. Let it be true for $n=k$, then we have \begin{align}1(1 !)+2(2 !)+3(3 !)+\ldots+k(k !)& ... to both sides of the induction hypothesis(i), we have \begin{align}1(1 !)+2(2 !)+3(3 !)+\ldots+k(k !)+
- Question 8 Exercise 7.1
- s+2^n 1=2^n-1$. ====Solution==== 1. For $n=1$, we have $1=2^1-1=1$. Thus it is true for $n=1$. 2. Let ... to both sides of the induction hypothesis (i), we have \begin{align}1+2+2^2+2^3+\ldots+2^{k-1}-2^k & =2^
- Question 5 and 6 Exercise 7.3
- and neglecting $x^2$ and higher powers of $x$, we have $$ \begin{aligned} & =1+\frac{5 x}{8}-\frac{5 x}{... lying and neglecting $\frac{1}{x^3}$ and so on we have $$ \begin{aligned} & =1 \frac{2}{x}+\frac{3}{x^2}
- Question 11 Exercise 7.3
- )}{3 !} x^3+\ldots$ Comparing both the series, we have $n x=\frac{1}{2^2}=\frac{1}{4}$ (1) and $\frac{n(... \cdot \frac{1}{2^4}$ Taking square of Eq.(1), we have $n^2 x^2=\frac{1}{16}$ Dividing Eq.(2) by Eq.(3),
- Question 7 & 8 Review Exercise 7
- he induction hypothesis. (3.) For $n=k+1$ then we have $$ \begin{aligned} & 7^{k+1}-3^{k+1}=7.7^k-3.3^k ... 1+k x) $$ 3. Now for $\boldsymbol{n}=k+1$ then we have $$ \begin{aligned} & (1+x)^{k+1}=(1+x)^k(1+x) \\
- Question 5 Exercise 7.1
- to both sides of the induction hypothesis (i), we have \begin{align}1^3+2^3+3^3+\ldots+k^3+(k+1)^3 & =[\
- Question 7 Exercise 7.1
- to both sides of the induction hypothesis (i), we have \begin{align}1.2+2.3+3.4+\ldots+k(k+1)+(k+1)(k+2)
- Question 11 Exercise 7.1
- 1}$ to both sides of the induction hypothesis, we have \begin{align}(\begin{array}{l} 2 \\ 2 \end{array}
- Question 12 Exercise 7.1
- b{Z}\cdots (i)\end{align} 3. For $n=k+1$, then we have \begin{align}\dfrac{10^{k+1+1}-9(k+1)-10}{81}& =\
- Question 13 Exercise 7.1
- $ then $k !>k^2\cdots(i)$ 3. For $n=k+1$ then we have \begin{align} & (k+1) !=(k+1) k !>(k+1)(k+1) \\ &