Find the coefficient of x2 in the expansion of e2x+3 as a series in powers of x.
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Solution
In the exponential series ex=1+x1!+x22!+x23!+... Hence e2x+3=e3.e2x=e3[1+x1!+x22!+x23!+...] Thus, the coefficient of x2 in the expansion of e2x+3 is e3.222!=2e3.