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- Question 10 Exercise 7.2
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... $. Solution: We know that $$ \left.(1+x)^n=\left(\begin{array}{l} n \\ \vdots \end{array}\right)+\left(\begin{array}{l} m \\ 1 \end{array}\right) x+\left(\begin{array}{l} n \\ 2 \end{array}\right) x^2-\ldots+
- Question 10 Exercise 7.1
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... an. =====Question 10===== Establish the formulas below by mathematical induction, $\left(\begin{array}{1}5 \\5 \end{array}\right)+\left(\begin{array}{l}6 \\ 5\end{array}\right)+\left(\begin{arra
- Question 11 Exercise 7.1
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. =====Question 11===== \begin{align} & \left(\begin{array}{l} 2 \\ 2 \end{array}\right)+\left(\begin{array}{l} 3 \\ 2 \end{array}\right)+\left(\begin{
- Question 7 Exercise 7.2
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... t{3})^5$ ====Solution==== Using binomial formula \begin{align}(2+\sqrt{3})^5+(2 \cdot \sqrt{3})^5& =[(... \cdot(\sqrt{3})^5\end{align} simplifing, we get \begin{align} & =2 \cdot 2^5+2^5 C_2 \cdot 2^3 \cdot(... 2})^{-}$ ====Solution==== Using binomial formula \begin{align} (1+\sqrt{2})^4-(1-\sqrt{2})^4 & =[1+{ }
- Question 1 Review Exercise 7
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... $2520$</collapse> ii. How many two digits odd numbers can be formed form the digits $\{1,2,3,4,5,6,7\}$ if repeated digits are allowed? * %%(a)%% $14... c): $28$ </collapse> iii. How many six digits number can be formed from the digits $\{1,2,3,4,6,7,8\}
- Question 3 Exercise 7.2
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... x^2}{3}$ and $b=-\dfrac{3}{2 x}$. Let $T_{r+1}$ be the term independent of $x$ in the given expansion. $T_{r+1}$ of the given expansion is: \begin{align}T_{r+1}&=\dfrac{9 !}{(9-r) ! r !}(\dfrac... rrow r=6 $$ Putting $r=6$ in the above $T_{r+1}$ \begin{align}T_{6-1}&=\dfrac{9 !}{(9-6) ! 6 !}\cdot \
- Question 4 Exercise 7.2
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... n $n=20, \quad a=x^2$ and $b=-x$. Let $T_{r, 1}$ be the term containing $x^{23}$ that is: \begin{align}T_{r-1}&=\dfrac{20 !}{(20-r) ! r !}(x^2)^{20 r}(-x... _{r-1}$ containing $x^{33}$ is possibic only if \begin{align}x^{40 \cdot r}&=x^{23}\\ \Rightarrow 40-
- Question 10 Exercise 7.3
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... tion: The given series is binomial series. Let it be identical with the expansion of $(1+x)^n$ that is $$ \begin{aligned} & 1+n x+\frac{n(n-1)}{2 !} x^2 \\ & +... rac{1}{16}$ Dividing Eq.(2) by Eq.(3), we get $$ \begin{aligned} & \frac{n-1}{2 n}=\frac{3}{32} \cdot
- Question 2 Exercise 7.3
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... laces. (i) $\sqrt{26}$ Solution: We are given $$ \begin{aligned} & \sqrt{26}=\sqrt{25+1} \\ & =\sqrt{2... }} \end{aligned} $$ Using binomial expansion $$ \begin{aligned} & \sqrt{26}=5\left[1+\frac{1}{25}\rig... } \text {. } $$ Using binomial expansion now $$ \begin{aligned} & =1-\frac{1}{2}(-0.002)+ \\ & \frac{
- Question 5 and 6 Exercise 7.3
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... 5 If $x$ is such that $x^2$ ard higher of $x$ may be negleeled. then show that $$ \frac{(8+3 x)^{\frac... 3}-3 x j^{\frac{2}{3}}}{2 \cdot 3 x+4-5 x} $$ $$ \begin{aligned} & =\frac{8^{\frac{2}{3}}\left(1+\frac... and neglecting $x^2$ and higher powers of $x$ $$ \begin{aligned} & =\left(1-\frac{3 x}{2}+\frac{x}{4}+
- Question 5 Exercise 7.2
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... $b=b x$ and $n=8$ Since $n-8$ is a the even number of terms in the expansion are $8+1=9$ The middle... (b x)^r$$ To get middle term $T_5$, we put $r=4$ \begin{align}T_5&=\dfrac{8 !}{(8-4) ! 4 !}(\dfrac{a}{... $ and $n=9$. Since $n=9$ is odd so the total number of terms in the expansion are $9+1=10$. So in t
- Question 12 Exercise 7.3
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... $ Now the above series is binomial series. Lel it be identical with the expansion of $(1+x)^n$ that is $$ \begin{aligned} & 10+n x+\frac{n(n-1)}{2 !} x^2+ \\ &... rac{1}{16}$ Dividing Eq.(2) by Eq.(3), we get $$ \begin{aligned} & \frac{n-1}{2 n}=\frac{1.3}{2 !} \cd
- Question 7 & 8 Review Exercise 7
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... des 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 ... ion hypothesis. (3.) For $n=k+1$ then we have $$ \begin{aligned} & 7^{k+1}-3^{k+1}=7.7^k-3.3^k \\ & =(... .3^k \\ & =4.7^k+3.7^k-3.3^k \end{aligned} $$ $$ \begin{aligned} & =4.7^k+3\left[7^k-3^k\right] \\ & \
- Question 12 Exercise 7.1
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... \mathbb{Z}$$ Thus it 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}&=\dfrac{5^{2 k+... s an integer. ====Solution==== 1. For $n=1$ then \begin{align}\dfrac{10^{n+1}-9 n-10}{81}&=\dfrac{10^{
- Question 9 Exercise 7.2
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... y-4$. Solution: When we put $x=12$, then the give becounes $$ \begin{aligned} & \left(x \quad y=20(12-y)^{20}\right. \\ & =12^{2 n}\left(\begin{array}{ll} 1 & \frac{y}{12} \end{array}\right)