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- Question 12 Exercise 7.1
- ===== Show by mathematical induction that $\dfrac{10^{n+1}-9 n-10}{81}$ is an integer. ====Solution==== 1. For $n=1$ then \begin{align}\dfrac{10^{n+1}-9 n-10}{81}&=\dfrac{10^{i+1}-9.1-10}{81} \\ & =\dfrac{100-9-10}{81}=1 \in \mathbb{Z}\end{align}
- Question 3 Exercise 7.2
- dent of $x$ in the expansion $(x-\dfrac{3}{x^4})^{10}$ ====Solution==== In the above expansion $n=10, \quad a=x$ and $b=-\dfrac{3}{x^4}$. Let $T_{r+1}$ b... given expansion is: \begin{align} T_{r+1}&=\dfrac{10 !}{(10-r) ! r !}(x)^{10 \cdot r}(-\dfrac{3}{x^4})^r \\ & =\dfrac{10 !}{(10-r) ! r !} \cdot(-3)^r x^{1
- Question 5 & 6 Review Exercise 7
- ion of $\left(\frac{2}{x^2}+\frac{x^2}{2}\right)^{10}$ ? Solution: Here $n=10, a^{\prime}=\frac{2}{x^2}$ and $b=\frac{x^2}{2}$. Let $T_{r+1}$ be the term ... f $x$ that is: $$ \begin{aligned} & T_{r+1}=\frac{10 !}{(10-r) ! r !}\left(\frac{2}{x^2}\right)^{10 r}\left(\frac{x^2}{2}\right)^r \\ & =\frac{10 !}{(10-r
- Question 5 Exercise 7.2
- e total number of terms in the expansion are $9+1=10$. So in this we have two middle terms that are $... ^4 \cdot \dfrac{1}{2^5} \cdot(x^2)^5\\ &=-\dfrac{5103}{16} x^{14} \end{align} Thus the two middle term... dfrac{15309}{8} x^{13} \text { and } T_6=-\dfrac{5103}{16} x^{14}$$ =====Question 5(iii)===== Find m... e term in the expansion of $(3 x^2-\dfrac{y}{3})^{10}$ ====Solution==== Since we see that $a=3 x^2$, $
- Question 8 Exercise 7.2
- Q8 Find numerically greatest term in $(3-2 x)^{10}$. when $x=\frac{3}{4}$. Solution: We first write in form $\left(3-2,1^{10}=3^{10}\left(1-\frac{3 x}{2}\right)^{10}\right.$. The numerically greatest term in $\left(1-\frac{3 x}{2}\ri
- Question 2 Exercise 7.2
- ext {th }}$ term in $(\dfrac{x}{2}-\dfrac{3}{y})^{10}$ ====Solution==== In the above $n=10$, $a=\dfrac{x}{2}$ and $b=-\dfrac{3}{y}$. Thus the general term of the given expansion is: $$T_{r+1}=\dfrac{10 !}{(10-r) ! r !}(\dfrac{x}{2})^{10-r}(-\dfrac{3}{y})^r$$ For $8^{\text {th }}$ term, putting $r=7$ \b
- Question 1 Exercise 7.2
- \dfrac{5}{2}}+5 \cdot y^2 \cdot y^{-\dfrac{1}{2}}+10 \cdot y^{\dfrac{3}{2}} \cdot y^{-1}\\ &+ 10 \cdot y \cdot y^{-\dfrac{3}{2}}+5 \cdot y^{\dfrac{1}{2}} ... 1}{2}}y^{\dfrac{1}{2}}+5 \cdot y^{2-\dfrac{1}{2}}+10 \cdot y^{\dfrac{3}{2}-\dfrac{1}{2}-\dfrac{1}{2}}\\ &+10 \cdot y^{1-\dfrac{3}{2}} +5 \cdot y^{\dfrac{1}{2}
- Question 1 Review Exercise 7
- er of terms in expansion of $(\sqrt{x}+\sqrt{y})^{10}+(\sqrt{x}+\sqrt{y})^{10}$ is (a) 6 (b) 11 (c) 20 (d) 5 (v) $(\sqrt{2}+1)^5+(\sqrt{2}-1)^5=$ (a) 58 (... ) $2^{30}$ (c) $2^{29}$ (d) $2^{28}$ (viii) ${ }^{10} C_1+{ }^{10} C_3+{ }^{10} C_5+\cdots+{ }^{10} C_9$ $=$ (a) 512 (b) 1024 (c) 2048 (d) 1023
- Question 10 Exercise 7.1
- ====== Question 10 Exercise 7.1 ====== Solutions of Question 10 of Exercise 7.1 of Unit 07: Permutation, Combination and... KPTB or KPTBB) Peshawar, Pakistan. =====Question 10===== Establish the formulas below by mathematical
- Question 10 Exercise 7.2
- ====== Question 10 Exercise 7.2 ====== Solutions of Question 10 of Exercise 7.2 of Unit 07: Permutation, Combination and... tbook Board (KPTB or KPTBB) Peshawar, Pakistan. Q10 Show that the sum of binomial coefticients of ord
- Question 10 Exercise 7.3
- ====== Question 10 Exercise 7.3 ====== Solutions of Question 10 of Exercise 7.3 of Unit 07: Permutation, Combination and... tbook Board (KPTB or KPTBB) Peshawar, Pakistan. Q10 Find the sum of the following series: (i) $1-\fra
- Question 9 Exercise 7.1
- align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-1-p10|Question 10 >]]</btn></text>
- Question 11 Exercise 7.1
- tn type="primary">[[math-11-kpk:sol:unit07:ex7-1-p10 |< Question 10 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-
- Question 7 Exercise 7.2
- cdot 2 \cdot(\sqrt{3})^4 \\ & =2 \cdot 32+2 \cdot 10 \cdot 8 \cdot 3+2 \cdot 5 \cdot 2 \cdot 9 \\ & =6... }{l} 5 \\ 4 \end{array}\right) a b^4\right] \\ & =10 a^4 b+20 a^2 b^3+2 a b^4 . \end{aligned} $$ ====G
- Question 9 Exercise 7.2
- align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-2-p10|Question 10 >]]</btn></text>