[tex](a + b)^{n} = \sum\limits_{i = 0}^{n}\binom{n}{k}a^{i}\cdot b^{n-i}[/tex][tex]\text{ Unde $\binom{n}{k}$ este al k-lea coeficient binomial din al n-lea rand din triunghiul lui Pascal.}[/tex][tex](2x + 1)^{4} = \sum\limits_{i=0}^{4} \binom{4}{i} (2x)^{i} \cdot 1^{4 - i} = \sum\limits_{i = 0}^{4} \binom{4}{i} 2^{i} x^{i} = \binom{4}{0} 2^{0} x^{0} + \binom{4}{1} 2^{1} x^{1} + \binom{4}{2} 2^{2} x^{2} + \binom{4}{3} 2^{3} x^{3} + \binom{4}{4} 2^{4} x^{4} = 1 + 8x + 24x^2 + 32x^3 + 16x^4[/tex]
[tex](2x-3)^{5} = \sum\limits_{i=0}^{5} \binom{5}{i}(2x)^{i} \cdot (-3)^{5 - i} = \binom{5}{0}(2x)^{0} \cdot (-3)^{5 - 0} + \binom{5}{1}(2x)^{1} \cdot (-3)^{5 - 1} + \binom{5}{2}(2x)^{2} \cdot (-3)^{5 - 2} + \binom{5}{3}(2x)^{3} \cdot (-3)^{5 - 3} + \binom{5}{4}(2x)^{4} \cdot (-3)^{5 - 4} + \binom{5}{5}(2x)^{5} \cdot (-3)^{5 - 5} = \\32x^{5} -240x^{4} + 720x^{3} - 1080x^{2} + 810x - 243[/tex]
Și ultimul îl faci singur.
