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- Question 1 Exercise 7.1
- Question 2 Exercise 7.1
- Question 3 Exercise 7.1
- Question 4 Exercise 7.1
- Question 5 Exercise 7.1
- Question 6 Exercise 7.1
- Question 7 Exercise 7.1
- Question 8 Exercise 7.1
- Question 9 Exercise 7.1
- Question 10 Exercise 7.1
- Question 11 Exercise 7.1
- Question 12 Exercise 7.1
- Question 13 Exercise 7.1
- Question 14 Exercise 7.1
- Question 15 Exercise 7.1
- Question 1 Exercise 7.2
- Question 2 Exercise 7.2
- Question 3 Exercise 7.2
- Question 4 Exercise 7.2
- Question 5 Exercise 7.2
- Question 6 Exercise 7.2
- Question 7 Exercise 7.2
- Question 8 Exercise 7.2
- Question 9 Exercise 7.2
- Question 10 Exercise 7.2
- Question 11 Exercise 7.2
- Question 1 Exercise 7.3
- Question 2 Exercise 7.3
- Question 3 Exercise 7.3
- Question 4 Exercise 7.3
- Question 5 and 6 Exercise 7.3
- Question 7 and 8 Exercise 7.3
- Question 9 Exercise 7.3
- Question 10 Exercise 7.3
- Question 11 Exercise 7.3
- Question 12 Exercise 7.3
- Question 13 Exercise 7.3
- Question 14 Exercise 7.3
- Question 1 Review Exercise 7
- Question 2 Review Exercise 7
- Question 3 & 4 Review Exercise 7
- Question 5 & 6 Review Exercise 7
- Question 7 & 8 Review Exercise 7
- Question 9 and 10 Review Exercise 7
- Question 11 Review Exercise 7
Fulltext results:
- Question 4 Exercise 7.1
- the formulas below by mathematical induction $3+7+11+\cdots+(4 n-1)=n(2 n+1)$ ====Solution==== 1. For ... et it be true for $n=k$, we have \begin{align}3+7+11+\cdots+(4 k-1) & =k(2 k+1)....(i) \end{align} 3.... duction hypothesis (i), we have \begin{align} 3+7+11+\cdots+(4 k-1)+[4(k+1)-1] & =k(2 k+1)+4(k+1)-1 \\... \\ & =(k+1)[2 k+2+1] \\ & =(k+1)[2(k+1)+1] \\ 3+7+11+\cdots+(4 k-1)+[4(k+ 1)-1]&=(k+1)[2(k+1)+1]\end{a
- Question 10 Exercise 7.3
- . (ii) $1+\frac{5}{8}+\frac{5.8}{8.12}+\frac{5.8 .11}{8.12 .16}+\ldots$ Solution: The given series is... }=1+\frac{5}{8}+\frac{5.8}{8.12} \\ & +\frac{5.8 .11}{8.12 .16}+\ldots \\ & \Rightarrow\left(\frac{24}... }=1+\frac{5}{8}+\frac{5.8}{8.12} \\ & +\frac{5.8 .11}{8.12 .16}+\ldots . \end{aligned} $$ Hence the s... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-3-p7 |< Question 9 ]]</btn></
- Question 11 Exercise 7.1
- ====== Question 11 Exercise 7.1 ====== Solutions of Question 11 of Exercise 7.1 of Unit 07: Permutation, Combination and... KPTB or KPTBB) Peshawar, Pakistan. =====Question 11===== \begin{align} & \left(\begin{array}{l} 2 \\ ... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-1-p10 |< Question 10 ]]</btn>
- Question 11 Exercise 7.3
- ====== Question 11 Exercise 7.3 ====== Solutions of Question 11 of Exercise 7.3 of Unit 07: Permutation, Combination and... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-3-p8 |< Question 10 ]]</btn><... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-3-p10|Question 12 >]]</btn></t
- Question 10 Exercise 7.1
- === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-1-p9 |< Question 9 ]]</btn></... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-1-p11|Question 11 >]]</btn></text>
- Question 12 Exercise 7.1
- === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-1-p11 |< Question 11 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-1-p13|Question 13 >]]</btn></t
- Question 5 Exercise 7.2
- number of terms in the expansion are $10_{\neg} 1=11$. The middle term is only one and is $(\dfrac{10... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-2-p4 |< Question 4 ]]</btn></... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-2-p6|Question 6 >]]</btn></tex
- Question 8 Exercise 7.2
- =\frac{3}{2} \frac{3}{2}: 8 \\ & \therefore \frac{11}{1} \cdot i \quad 7 \\ & =5.82 \quad \because \qu... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-2-p7 |< Question 7 ]]</btn></... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-2-p9|Question 9 >]]</btn></tex
- Question 10 Exercise 7.2
- === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-2-p9 |< Question 9 ]]</btn></... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-2-p11|Question 11 >]]</btn></text>
- Question 11 Exercise 7.2
- ====== Question 11 Exercise 7.2 ====== Solutions of Question 11 of Exercise 7.2 of Unit 07: Permutation, Combination and... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-2-p10 |< Question 10 ]]</btn>
- Question 7 and 8 Exercise 7.3
- a-b x^2$ and tve oltain: $\left.(1+x) \frac{1}{4}+11-x\right) \frac{1}{4}=2-\frac{3 x^2}{16}$. From th... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-3-p5 |< Question 5 & 6 ]]</bt... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-3-p7|Question 9 >]]</btn></tex
- Question 12 Exercise 7.3
- === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-3-p9 |< Question 11 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-3-p11|Question 13 >]]</btn></t
- Question 9 and 10 Review Exercise 7
- === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:Re-ex7-p5 |< Question 7 & 8 ]]</bt... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit07:Re-ex7-p7|Question 11 >]]</btn></text>
- Question 11 Review Exercise 7
- ====== Question 11 Review Exercise 7 ====== Solutions of Question 11 of Review Exercise 7 of Unit 07: Permutation, Combi... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:Re-ex7-p6 |< Question 9 & 10 ]]</b
- Question 2 Exercise 7.1
- === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-1-p1 |< Question 1 ]]</btn></... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-1-p3|Question 3 >]]</btn></tex