Using binomial expansion express (1.2)⁵

1.	Using binomial expansion express (1.2)⁵
Answer» (1.2)⁵ = (1 + 0.2)⁵ = (1 + \(\frac{2}{10}\))⁵ = (1 + \(\frac{1}{5}\))⁵ = ⁵C₀1⁵ + ⁵C₁1⁴ (\(\frac{1}{5}\)) + ⁵C₂1³( \(\frac{1}{5}\))² + ⁵C₃1² (\(\frac{1}{5}\))³ + ⁵C₄1 x (\(\frac{1}{5}\))⁴ + ⁵C₅ (\(\frac{1}{5}\))⁴ + ⁵C₅ (\(\frac{1}{5}\))⁵ = 1 + 5 x \(\frac{1}{5}\)+ \(\frac{5\times 4}{2\times 1}\) x \(\frac{1}{25}\) + \(\frac{5\times 4}{2\times 1}\) x \(\frac{1}{125}\) + 5 x \(\frac{1}{5}\) x (0.2)³ + (0.2)⁵ = 1 + 1 + 2 x 0.2 + 2 x (0.2)² + (0.2)³ + (0.2)⁵ = 2 + 0.4 + 2 x 0.04 + 0.008 + 0.00032 = 2.4 + 0.08 + 0.00832 = 2.48832

Answer»

(1.2)⁵ = (1 + 0.2)⁵

= (1 + \(\frac{2}{10}\))⁵

= (1 + \(\frac{1}{5}\))⁵

= ⁵C₀1⁵ + ⁵C₁1⁴ (\(\frac{1}{5}\)) + ⁵C₂1³( \(\frac{1}{5}\))² + ⁵C₃1² (\(\frac{1}{5}\))³ + ⁵C₄1 x (\(\frac{1}{5}\))⁴ + ⁵C₅ (\(\frac{1}{5}\))⁴ + ⁵C₅ (\(\frac{1}{5}\))⁵

= 1 + 5 x \(\frac{1}{5}\)+ \(\frac{5\times 4}{2\times 1}\) x \(\frac{1}{25}\) + \(\frac{5\times 4}{2\times 1}\) x \(\frac{1}{125}\) + 5 x \(\frac{1}{5}\) x (0.2)³ + (0.2)⁵

= 1 + 1 + 2 x 0.2 + 2 x (0.2)² + (0.2)³ + (0.2)⁵

= 2 + 0.4 + 2 x 0.04 + 0.008 + 0.00032

= 2.4 + 0.08 + 0.00832

= 2.48832

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