Step 1: \((a+b)^7 = \sum_{r=0}^{7} \binom{7}{r} a^{7-r} b^r\)
Step 2: \(= \binom{7}{0}a^7 + \binom{7}{1}a^6b + \binom{7}{2}a^5b^2 + \binom{7}{3}a^4b^3 + \binom{7}{4}a^3b^4 + \binom{7}{5}a^2b^5 + \binom{7}{6}ab^6 + \binom{7}{7}b^7\)
Step 3: \(= a^7 + 7a^6b + 21a^5b^2 + 35a^4b^3 + 35a^3b^4 + 21a^2b^5 + 7ab^6 + b^7\)

Step 1: \((2x-3y)^4 = \sum_{r=0}^{4} \binom{4}{r} (2x)^{4-r} (-3y)^r\)
Step 2: \(= \binom{4}{0}(2x)^4 + \binom{4}{1}(2x)^3(-3y) + \binom{4}{2}(2x)^2(9y^2) + \binom{4}{3}(2x)(-27y^3) + \binom{4}{4}(81y^4)\)
Step 3: \(= 16x^4 + 4 \cdot 8x^3 \cdot (-3y) + 6 \cdot 4x^2 \cdot 9y^2 + 4 \cdot 2x \cdot (-27y^3) + 81y^4\)
Step 4: \(= 16x^4 – 96x^3y + 216x^2y^2 – 216xy^3 + 81y^4\)

Step 1: \(= \binom{5}{0}(2x)^5 + \binom{5}{1}(2x)^4 y^2 + \binom{5}{2}(2x)^3 (y^2)^2 + \binom{5}{3}(2x)^2 (y^2)^3 + \binom{5}{4}(2x)(y^2)^4 + \binom{5}{5}(y^2)^5\)
Step 2: \(= 1 \cdot 32x^5 + 5 \cdot 16x^4 \cdot y^2 + 10 \cdot 8x^3 \cdot y^4 + 10 \cdot 4x^2 \cdot y^6 + 5 \cdot 2x \cdot y^8 + 1 \cdot y^{10}\)
Step 3: \(= 32x^5 + 80x^4y^2 + 80x^3y^4 + 40x^2y^6 + 10xy^8 + y^{10}\)

Step 1: \(\sum_{r=0}^{5} \binom{5}{r} \left(\frac{x}{2}\right)^{5-r} \left(\frac{2}{y}\right)^r\)
Step 2: \(= \binom{5}{0}\frac{x^5}{32} + \binom{5}{1}\frac{x^4}{16}\cdot\frac{2}{y} + \binom{5}{2}\frac{x^3}{8}\cdot\frac{4}{y^2} + \binom{5}{3}\frac{x^2}{4}\cdot\frac{8}{y^3} + \binom{5}{4}\frac{x}{2}\cdot\frac{16}{y^4} + \binom{5}{5}\frac{32}{y^5}\)
Step 3: Simplify: \(\frac{x^5}{32} + \frac{5x^4}{8y} + \frac{10x^3}{y^2?}\) Wait recalc: \(\binom{5}{3}=10\), \(\frac{x^2}{4}\cdot\frac{8}{y^3}= \frac{2x^2}{y^3}\), so term = \(20x^2/y^3\)
Final: \(\frac{x^5}{32} + \frac{5x^4}{8y} + \frac{5x^3}{y^2} + \frac{20x^2}{y^3} + \frac{40x}{y^4} + \frac{32}{y^5}\)

Step 1: \(t_{r+1} = \binom{5}{r} (x^2)^{5-r} \left(-\frac{1}{x}\right)^r = \binom{5}{r} (-1)^r x^{10-3r}\)
Step 2: For \(r=0\): \(x^{10}\), \(r=1\): \(-5x^7\), \(r=2\): \(10x^4\), \(r=3\): \(-10x\), \(r=4\): \(5x^{-2}\), \(r=5\): \(-x^{-5}\)
Final: \(x^{10} – 5x^7 + 10x^4 – 10x + \frac{5}{x^2} – \frac{1}{x^5}\)

Step 1: General term: \(t_{r+1} = \binom{12}{r} (2x)^{12-r} y^r\)
Step 2: 7th term ⇒ \(r+1=7\) ⇒ \(r=6\)
Step 3: \(t_7 = \binom{12}{6} (2x)^6 y^6 = 924 \cdot 64 x^6 y^6 = 59136 \, x^6 y^6\)

Step 1: \(t_{r+1} = \binom{8}{r} (2x^2)^{8-r} (x^{-1})^r = \binom{8}{r} 2^{8-r} x^{16-3r}\)
Step 2: 4th term ⇒ \(r=3\)
Step 3: \(t_4 = \binom{8}{3} 2^{5} x^{16-9} = 56 \times 32 \times x^7 = 1792 x^7\)

Step 1: \(t_{r+1} = \binom{7}{r} x^{7-r} \left(-\frac{2}{x}\right)^r = \binom{7}{r} (-2)^r x^{7-2r}\)
Step 2: 5th term ⇒ \(r=4\)
Step 3: \(t_5 = \binom{7}{4} (-2)^4 x^{7-8} = 35 \times 16 \times x^{-1} = \frac{560}{x}\)

\(t_{r+1} = \binom{6}{r} (x^2)^{6-r} (x^{-1})^r = \binom{6}{r} x^{12-3r}\)

\(t_{r+1} = \binom{2n+1}{r} \left(\frac{a}{b}\right)^{2n+1-r} \left(\frac{b}{a}\right)^r = \binom{2n+1}{r} \left(\frac{a}{b}\right)^{2n+1-2r}\)

Step 1: \(t_{r+1} = \binom{10}{r} (x^2)^{10-r} \left(\frac{a}{x}\right)^r = \binom{10}{r} a^r x^{20-3r}\)
Step 2: For \(x^2\): \(20-3r=2\) ⇒ \(3r=18\) ⇒ \(r=6\)
Step 3: Coefficient = \(\binom{10}{6} a^6 = 210 a^6\)

Step 1: \(t_{r+1} = \binom{9}{r} (3x^2)^{9-r} \left(-\frac{1}{3x}\right)^r = \binom{9}{r} (-1)^r 3^{9-2r} x^{18-3r}\)
Step 2: For \(x^6\): \(18-3r=6\) ⇒ \(r=4\)
Step 3: Coefficient = \(\binom{9}{4} (-1)^4 3^{1} = 126 \times 3 = 378\)

Step 1: \(t_{r+1} = \binom{12}{r} x^{24-2r} \cdot x^{-r} = \binom{12}{r} x^{24-3r}\)
Step 2: Independent ⇒ \(24-3r=0\) ⇒ \(r=8\)
Step 3: Constant term = \(\binom{12}{8} = 495\)

Step 1: \(t_{r+1} = \binom{9}{r} \left(\frac{3x^2}{2}\right)^{9-r} \left(-\frac{1}{3x}\right)^r\)
Step 2: \(= \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} (-1)^r 3^{-r} x^{18-3r}\)
Step 3: Constant ⇒ \(18-3r=0\) ⇒ \(r=6\)
Step 4: Term = \(\binom{9}{6} \left(\frac{3}{2}\right)^{3} 3^{-6} = 84 \cdot \frac{27}{8} \cdot \frac{1}{729} = \frac{84}{216} = \frac{7}{18}\)

\(\left(1 + \frac{1}{x}\right)^{2n} = \sum_{r=0}^{2n} \binom{2n}{r} x^{-r}\)
Constant when \(r=0\) ⇒ term = \(1\)
Alternative: For \((x + \frac{1}{x})^{2n}\), constant term = \(\binom{2n}{n}\)

Step 1: 4th term = \(\binom{n}{3} (px)^{n-3} (1/x)^3 = \binom{n}{3} p^{n-3} x^{n-6}\)
Step 2: Independent ⇒ \(n-6=0\) ⇒ \(n=6\)
Step 3: \(t_4 = \binom{6}{3} p^{3} = 20p^3 = \frac{5}{2}\) ⇒ \(p^3 = \frac{1}{8}\) ⇒ \(p = \frac{1}{2}\)

Step 1: \((x+k)^5 = \sum_{r=0}^{5} \binom{5}{r} x^{5-r} k^r\)
Step 2: For \(x^2\): \(5-r=2\) ⇒ \(r=3\)
Step 3: Coefficient = \(\binom{5}{3} k^3 = 10k^3 = 90\) ⇒ \(k^3 = 9\) ⇒ \(k = \sqrt[3]{9}\)

\(\left(2 + \frac{1}{x}\right)^{17}\) ⇒ odd index \(n=17\) ⇒ two middle terms: \(t_9\) and \(t_{10}\)
\(t_9 = \binom{17}{8} 2^{9} x^{-8}\), \(t_{10} = \binom{17}{9} 2^{8} x^{-9}\)

Step 1: Middle term = \(t_{n+1} = \binom{2n}{n} x^n\)
Step 2: \(\binom{2n}{n} = \frac{(2n)!}{n!n!} = \frac{2^n (1\cdot3\cdot5\cdots(2n-1))}{n!}\)
Step 3: Hence \(t_{n+1} = \frac{1\cdot3\cdot5\cdots(2n-1)}{n!} 2^n x^n\)

\(\binom{2n+1}{r} = \binom{2n+1}{r+1}\)
For binomial coefficients: \(r + (r+1) = 2n+1\) ⇒ \(2r+1 = 2n+1\) ⇒ \(r = n\)

Step 1: Middle term of \((1+x)^{2n}\): \(\binom{2n}{n}\)
Step 2: Middle terms of \((1+x)^{2n-1}\): \(\binom{2n-1}{n-1}\) and \(\binom{2n-1}{n}\)
Step 3: Sum = \(\binom{2n-1}{n-1} + \binom{2n-1}{n} = \binom{2n}{n}\) (Pascal’s identity)

Step 1: Middle term is the \((n+1)\)th term
Step 2: \(t_{n+1} = \binom{2n}{n} (1)^{2n-n} x^n = \binom{2n}{n} x^n\)
Step 3: \(\binom{2n}{n} = \frac{(2n)!}{n!n!}\)