Mathematics

use the expansion (a+b)^4 to evaluate (1.03)^4 correct to 2 decimal places. use the expansion (a+b)^4 to evaluate (1.03)^4 correct to 2 decimal places. @Mathematics

OpenStudy (zarkon):

$(a+b)^n=\sum_{k=0}^{n}{n\choose k}a^kb^{n-k}$ $(1+.03)^4=\sum_{k=0}^{4}{4\choose k}(.03)^{n-k}$

OpenStudy (zarkon):

$(1+.03)^4=\sum_{k=0}^{4}{4\choose k}(.03)^{4-k}$

OpenStudy (anonymous):

Let a = 1, b = 0.03 (1+0.03)^4 = 1+0.03*4+0.03^2*6+0.03^3*4+0.03^4 Only the first 2 terms are important here since you need 2 d.p. $(1+0.03)^4 \approx 1+0.03*4 = 1.12$

OpenStudy (anonymous):

No wait! 0.03^2*6=0.0054 so your answer is 1.13