All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
e^x (1+ \mathrm{sin}2x)

= (1 + x + \frac{x^2}{2!} + \ldots)(1+2x-\ldots)

= 1 + 3x + \frac{5}{2}x^2 + \ldots

(ii)
(1 + \frac{4}{3}x)^n

= 1 + n\frac{4}{3}x + \frac{n(n-1)}{2!}(\frac{4}{3}x)^2 + \ldots

Since the first two terms are equal,

\frac{4}{3}n = 3 \Rightarrow n = \frac{9}{4}

Third term = \frac{n(n-1)}{2!}(\frac{4}{3}x)^2 = \frac{5}{2}x^2

KS Comments:

Students should know how to find coefficient of terms efficiently. Refer here if they don’t. This is rather simple since they only expect students to do up to 3 terms.

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