1. **Evaluate binomial coefficients:** (i) $\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21$ (ii) $\binom{8}{0} = 1$ (choosing zero items always 1 way) (iii) $\binom{7}{4} = \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$ (iv) $\binom{12}{7} = \binom{12}{5} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792$ (v) $\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$ (vi) $\binom{6}{4} = \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$ (vii) $\binom{7}{1} = 7$ (viii) $\binom{8}{8} = 1$ 2. **Expand binomials using Binomial Theorem:** Formula: $$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$$ (i) $(4 + 3x)^5$: Calculate terms: $\binom{5}{0}4^5(3x)^0 = 1024$ $\binom{5}{1}4^4(3x)^1 = 5 \times 256 \times 3x = 3840x$ $\binom{5}{2}4^3(3x)^2 = 10 \times 64 \times 9x^2 = 5760x^2$ $\binom{5}{3}4^2(3x)^3 = 10 \times 16 \times 27x^3 = 4320x^3$ $\binom{5}{4}4^1(3x)^4 = 5 \times 4 \times 81x^4 = 1620x^4$ $\binom{5}{5}4^0(3x)^5 = 1 \times 1 \times 243x^5 = 243x^5$ Expansion: $$1024 + 3840x + 5760x^2 + 4320x^3 + 1620x^4 + 243x^5$$ (ii) $(2x + 3y)^4$: Terms: $\binom{4}{0}(2x)^4(3y)^0 = 16x^4$ $\binom{4}{1}(2x)^3(3y)^1 = 4 \times 8x^3 \times 3y = 96x^3y$ $\binom{4}{2}(2x)^2(3y)^2 = 6 \times 4x^2 \times 9y^2 = 216x^2y^2$ $\binom{4}{3}(2x)^1(3y)^3 = 4 \times 2x \times 27y^3 = 216xy^3$ $\binom{4}{4}(2x)^0(3y)^4 = 1 \times 1 \times 81y^4 = 81y^4$ Expansion: $$16x^4 + 96x^3y + 216x^2y^2 + 216xy^3 + 81y^4$$ (iii) $(2p - \frac{1}{2}q)^5$: Use $a=2p$, $b=-\frac{1}{2}q$ Calculate terms: $\binom{5}{0}(2p)^5(-\frac{1}{2}q)^0 = 32p^5$ $\binom{5}{1}(2p)^4(-\frac{1}{2}q)^1 = 5 \times 16p^4 \times (-\frac{1}{2}q) = -40p^4q$ $\binom{5}{2}(2p)^3(-\frac{1}{2}q)^2 = 10 \times 8p^3 \times \frac{1}{4}q^2 = 20p^3q^2$ $\binom{5}{3}(2p)^2(-\frac{1}{2}q)^3 = 10 \times 4p^2 \times (-\frac{1}{8}q^3) = -5p^2q^3$ $\binom{5}{4}(2p)^1(-\frac{1}{2}q)^4 = 5 \times 2p \times \frac{1}{16}q^4 = \frac{5}{8}pq^4$ $\binom{5}{5}(2p)^0(-\frac{1}{2}q)^5 = 1 \times 1 \times (-\frac{1}{32}q^5) = -\frac{1}{32}q^5$ Expansion: $$32p^5 - 40p^4q + 20p^3q^2 - 5p^2q^3 + \frac{5}{8}pq^4 - \frac{1}{32}q^5$$ (iv) $(x + \frac{3}{x})^5$: Use $a=x$, $b=\frac{3}{x}$ Calculate terms: $\binom{5}{0}x^5(\frac{3}{x})^0 = x^5$ $\binom{5}{1}x^4(\frac{3}{x})^1 = 5x^4 \times \frac{3}{x} = 15x^3$ $\binom{5}{2}x^3(\frac{3}{x})^2 = 10x^3 \times \frac{9}{x^2} = 90x$ $\binom{5}{3}x^2(\frac{3}{x})^3 = 10x^2 \times \frac{27}{x^3} = 270\frac{1}{x}$ $\binom{5}{4}x^1(\frac{3}{x})^4 = 5x \times \frac{81}{x^4} = 405\frac{1}{x^3}$ $\binom{5}{5}x^0(\frac{3}{x})^5 = 1 \times \frac{243}{x^5} = \frac{243}{x^5}$ Expansion: $$x^5 + 15x^3 + 90x + 270\frac{1}{x} + 405\frac{1}{x^3} + \frac{243}{x^5}$$ (v) $(4x^2 - \frac{3}{x^2})^4$: Use $a=4x^2$, $b=-\frac{3}{x^2}$ Calculate terms: $\binom{4}{0}(4x^2)^4(-\frac{3}{x^2})^0 = 256x^8$ $\binom{4}{1}(4x^2)^3(-\frac{3}{x^2})^1 = 4 \times 64x^6 \times (-\frac{3}{x^2}) = -768x^4$ $\binom{4}{2}(4x^2)^2(-\frac{3}{x^2})^2 = 6 \times 16x^4 \times \frac{9}{x^4} = 864$ $\binom{4}{3}(4x^2)^1(-\frac{3}{x^2})^3 = 4 \times 4x^2 \times (-\frac{27}{x^6}) = -432\frac{1}{x^4}$ $\binom{4}{4}(4x^2)^0(-\frac{3}{x^2})^4 = 1 \times 1 \times \frac{81}{x^8} = \frac{81}{x^8}$ Expansion: $$256x^8 - 768x^4 + 864 - 432\frac{1}{x^4} + \frac{81}{x^8}$$ (vi) $(\frac{1}{3}x + \frac{2}{x^4})^5$: Use $a=\frac{1}{3}x$, $b=\frac{2}{x^4}$ Calculate terms: $\binom{5}{0}(\frac{1}{3}x)^5(\frac{2}{x^4})^0 = \frac{1}{243}x^5$ $\binom{5}{1}(\frac{1}{3}x)^4(\frac{2}{x^4})^1 = 5 \times \frac{1}{81}x^4 \times \frac{2}{x^4} = \frac{10}{81}$ $\binom{5}{2}(\frac{1}{3}x)^3(\frac{2}{x^4})^2 = 10 \times \frac{1}{27}x^3 \times \frac{4}{x^8} = \frac{40}{27}x^{-5}$ $\binom{5}{3}(\frac{1}{3}x)^2(\frac{2}{x^4})^3 = 10 \times \frac{1}{9}x^2 \times \frac{8}{x^{12}} = \frac{80}{9}x^{-10}$ $\binom{5}{4}(\frac{1}{3}x)^1(\frac{2}{x^4})^4 = 5 \times \frac{1}{3}x \times \frac{16}{x^{16}} = \frac{80}{3}x^{-15}$ $\binom{5}{5}(\frac{1}{3}x)^0(\frac{2}{x^4})^5 = 1 \times 1 \times \frac{32}{x^{20}} = 32x^{-20}$ Expansion: $$\frac{1}{243}x^5 + \frac{10}{81} + \frac{40}{27}x^{-5} + \frac{80}{9}x^{-10} + \frac{80}{3}x^{-15} + 32x^{-20}$$ 3. **First 4 terms in ascending powers of x:** (i) $(3 + 4x)^5$: Use $a=3$, $b=4x$ Terms for $k=0$ to $3$: $\binom{5}{0}3^5(4x)^0 = 243$ $\binom{5}{1}3^4(4x)^1 = 5 \times 81 \times 4x = 1620x$ $\binom{5}{2}3^3(4x)^2 = 10 \times 27 \times 16x^2 = 4320x^2$ $\binom{5}{3}3^2(4x)^3 = 10 \times 9 \times 64x^3 = 5760x^3$ First 4 terms: $$243 + 1620x + 4320x^2 + 5760x^3$$ (ii) $(2 - \frac{x}{4})^8$: $a=2$, $b=-\frac{x}{4}$ Terms $k=0$ to $3$: $\binom{8}{0}2^8(-\frac{x}{4})^0 = 256$ $\binom{8}{1}2^7(-\frac{x}{4})^1 = 8 \times 128 \times (-\frac{x}{4}) = -256x$ $\binom{8}{2}2^6(-\frac{x}{4})^2 = 28 \times 64 \times \frac{x^2}{16} = 112x^2$ $\binom{8}{3}2^5(-\frac{x}{4})^3 = 56 \times 32 \times (-\frac{x^3}{64}) = -28x^3$ First 4 terms: $$256 - 256x + 112x^2 - 28x^3$$ (iii) $(2p + \frac{1}{3}q)^7$: $a=2p$, $b=\frac{1}{3}q$ Terms $k=0$ to $3$: $\binom{7}{0}(2p)^7(\frac{1}{3}q)^0 = 128p^7$ $\binom{7}{1}(2p)^6(\frac{1}{3}q)^1 = 7 \times 64p^6 \times \frac{1}{3}q = \frac{448}{3}p^6q$ $\binom{7}{2}(2p)^5(\frac{1}{3}q)^2 = 21 \times 32p^5 \times \frac{1}{9}q^2 = \frac{224}{3}p^5q^2$ $\binom{7}{3}(2p)^4(\frac{1}{3}q)^3 = 35 \times 16p^4 \times \frac{1}{27}q^3 = \frac{560}{27}p^4q^3$ First 4 terms: $$128p^7 + \frac{448}{3}p^6q + \frac{224}{3}p^5q^2 + \frac{560}{27}p^4q^3$$ (iv) $(2a + 3b)^6$: $a=2a$, $b=3b$ Terms $k=0$ to $3$: $\binom{6}{0}(2a)^6(3b)^0 = 64a^6$ $\binom{6}{1}(2a)^5(3b)^1 = 6 \times 32a^5 \times 3b = 576a^5b$ $\binom{6}{2}(2a)^4(3b)^2 = 15 \times 16a^4 \times 9b^2 = 2160a^4b^2$ $\binom{6}{3}(2a)^3(3b)^3 = 20 \times 8a^3 \times 27b^3 = 4320a^3b^3$ First 4 terms: $$64a^6 + 576a^5b + 2160a^4b^2 + 4320a^3b^3$$ (v) $(\frac{1}{3} - 3x)^7$: $a=\frac{1}{3}$, $b=-3x$ Terms $k=0$ to $3$: $\binom{7}{0}(\frac{1}{3})^7(-3x)^0 = \frac{1}{2187}$ $\binom{7}{1}(\frac{1}{3})^6(-3x)^1 = 7 \times \frac{1}{729} \times (-3x) = -\frac{7}{243}x$ $\binom{7}{2}(\frac{1}{3})^5(-3x)^2 = 21 \times \frac{1}{243} \times 9x^2 = \frac{7}{9}x^2$ $\binom{7}{3}(\frac{1}{3})^4(-3x)^3 = 35 \times \frac{1}{81} \times (-27x^3) = -\frac{35}{9}x^3$ First 4 terms: $$\frac{1}{2187} - \frac{7}{243}x + \frac{7}{9}x^2 - \frac{35}{9}x^3$$ (vi) $(x - \frac{1}{x})^5$: $a=x$, $b=-\frac{1}{x}$ Terms $k=0$ to $3$: $\binom{5}{0}x^5(-\frac{1}{x})^0 = x^5$ $\binom{5}{1}x^4(-\frac{1}{x})^1 = 5x^4 \times (-\frac{1}{x}) = -5x^3$ $\binom{5}{2}x^3(-\frac{1}{x})^2 = 10x^3 \times \frac{1}{x^2} = 10x$ $\binom{5}{3}x^2(-\frac{1}{x})^3 = 10x^2 \times (-\frac{1}{x^3}) = -10\frac{1}{x}$ First 4 terms: $$x^5 - 5x^3 + 10x - 10\frac{1}{x}$$ 4. **First three terms in descending powers of x of $(3x - \frac{2}{x})^7$:** $a=3x$, $b=-\frac{2}{x}$ Terms $k=0$ to $2$: $\binom{7}{0}(3x)^7(-\frac{2}{x})^0 = 3^7 x^7 = 2187x^7$ $\binom{7}{1}(3x)^6(-\frac{2}{x})^1 = 7 \times 729x^6 \times (-\frac{2}{x}) = -10206x^5$ $\binom{7}{2}(3x)^5(-\frac{2}{x})^2 = 21 \times 243x^5 \times \frac{4}{x^2} = 20412x^3$ First 3 terms: $$2187x^7 - 10206x^5 + 20412x^3$$ 5. **First four terms in ascending powers of a of $(2a + \frac{1}{3}b)^6$:** $a=2a$, $b=\frac{1}{3}b$ Terms $k=0$ to $3$: $\binom{6}{0}(2a)^0(\frac{1}{3}b)^6 = \frac{1}{729}b^6$ $\binom{6}{1}(2a)^1(\frac{1}{3}b)^5 = 6 \times 2a \times \frac{1}{243}b^5 = \frac{12}{243}ab^5 = \frac{4}{81}ab^5$ $\binom{6}{2}(2a)^2(\frac{1}{3}b)^4 = 15 \times 4a^2 \times \frac{1}{81}b^4 = \frac{60}{81}a^2b^4 = \frac{20}{27}a^2b^4$ $\binom{6}{3}(2a)^3(\frac{1}{3}b)^3 = 20 \times 8a^3 \times \frac{1}{27}b^3 = \frac{160}{27}a^3b^3$ First 4 terms: $$\frac{1}{729}b^6 + \frac{4}{81}ab^5 + \frac{20}{27}a^2b^4 + \frac{160}{27}a^3b^3$$ 6. **Coefficient of $x^3$ in $(2x^2 - \frac{2}{x})^6$:** $a=2x^2$, $b=-\frac{2}{x}$ General term: $$T_{k+1} = \binom{6}{k} (2x^2)^{6-k} \left(-\frac{2}{x}\right)^k = \binom{6}{k} 2^{6-k} x^{2(6-k)} (-1)^k 2^k x^{-k} = \binom{6}{k} 2^6 (-1)^k x^{12 - 2k - k} = \binom{6}{k} 64 (-1)^k x^{12 - 3k}$$ We want power of $x^3$: $$12 - 3k = 3 \implies 3k = 9 \implies k=3$$ Coefficient: $$\binom{6}{3} 64 (-1)^3 = 20 \times 64 \times (-1) = -1280$$ 7. (i) **First three terms in descending powers of x of $(x + \frac{3}{x})^5$:** $a=x$, $b=\frac{3}{x}$ Terms $k=0$ to $2$: $\binom{5}{0}x^5(\frac{3}{x})^0 = x^5$ $\binom{5}{1}x^4(\frac{3}{x})^1 = 5x^4 \times \frac{3}{x} = 15x^3$ $\binom{5}{2}x^3(\frac{3}{x})^2 = 10x^3 \times \frac{9}{x^2} = 90x$ First 3 terms: $$x^5 + 15x^3 + 90x$$ (ii) **Coefficient of $x$ in $(2 + \frac{1}{x^2})(x + \frac{3}{x})^5$:** From above expansion: $(x + \frac{3}{x})^5 = x^5 + 15x^3 + 90x + \ldots$ Multiply by $(2 + \frac{1}{x^2})$: $$2(x^5 + 15x^3 + 90x) + \frac{1}{x^2}(x^5 + 15x^3 + 90x)$$ Terms contributing to $x$ power: From $2 \times 90x = 180x$ From $\frac{1}{x^2} \times 15x^3 = 15x$ Sum coefficient of $x$: $$180 + 15 = 195$$ 8. **In $(3 + px)^n$, first three terms are $243 + 810x + qx^2$:** General terms: $$T_1 = \binom{n}{0}3^n p^0 x^0 = 3^n = 243$$ So, $3^n = 243 \implies n=5$ (since $3^5=243$) $$T_2 = \binom{5}{1}3^{4} p x = 5 \times 81 \times p x = 405p x = 810x$$ Equate coefficients: $$405p = 810 \implies p=2$$ $$T_3 = \binom{5}{2}3^{3} p^2 x^2 = 10 \times 27 \times 4 x^2 = 1080 x^2$$ So, $q=1080$ 9. **Coefficient of middle term in $(2 + 3x)^6$:** Number of terms = $6+1=7$, middle term is term 4 ($k=3$) Coefficient: $$\binom{6}{3} 2^{6-3} 3^3 = 20 \times 8 \times 27 = 4320$$ **Final answers:** 1. (i) 21 (ii) 1 (iii) 35 (iv) 792 (v) 220 (vi) 15 (vii) 7 (viii) 1 2. Expansions as above 3. First 4 terms as above 4. First 3 terms as above 5. First 4 terms as above 6. Coefficient = -1280 7. (i) First 3 terms as above (ii) Coefficient of $x$ = 195 8. $n=5$, $p=2$, $q=1080$ 9. Coefficient middle term = 4320