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OpenStudy (asadkarim7):

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

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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

OpenStudy (asadkarim7):

Thank you all ov u guys <3 :D

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