(a) Write down the first four terms of the binomial expansion of ((2 – frac{1}{2})^{5}) in ascending powers of x.
(b) Use your expansion in (a) above to find, correct to two decimal places, the value of ((1.99)^{5}).
Explanation
(a) From Paschal’s triangle, coefficients of the power of 5 are
1, 5, 10, 10, 5, 1
((2 – frac{1}{2}x)^{5} = (1)(2^{5}) + 5(2^{4})(-frac{1}{2}x) + 10(2^{3})(-frac{1}{2}x)^{2} + 10(2^{2})(-frac{1}{2}x)^{3} + …)
= (32 – 40x + 20x^{2} – 5x^{3} + …)
(b) ((1.99)^{5} = (2 – 0.01)^{5} = (2 – frac{1}{2}(0.02))^{5})
Using the expansion above, (32 – 40x + 20x^{2} – 5x^{3} + …)
((1.99)^{5} approxeq 32 – 40(0.02) + 20(0.02)^{2} – 5(0.02)^{3} )
(approxeq 32 – 0.08 + 0.008)
= (31.928)
= (31.93) (to 2 decimal place)