Question 1

Find the sum of the following series;
$$\frac{2}{1!}+\frac{3}{2!}+\frac{6}{3!}+\frac{11}{4!}+\frac{18}{5!}+...$$

Solution

$$\frac{2}{1!}+\frac{3}{2!}+\frac{6}{3!}+\frac{11}{4!}+\frac{18}{5!}+...$$ = 2 + 1.5 + 1 + 0.5 + 0.15 + smaller insignificant values $$\approx$$ 5.2

Option 'A' = 3e - 1 = 3*2.718-1 = 7.15 therefore cannot be our answer

Option 'B' = 3(e - 1) = 3*(2.718-1) = 5.154 therefore can be our answer

Option 'C' = 3(e + 1) = 3*(2.718+1) = 11.15 therefore cannot be our answer

Option 'D' = 3e + 1 = 9.15 therefore cannot be our answer

Therefore option 'B' is our answer


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