Question 97

The sum of the series is:
$$\frac{1}{1.2.3}+\frac{1}{3.4.5}+\frac{1}{5.6.7}+...$$

Solution

$$\frac{1}{1.2.3}+\frac{1}{3.4.5}+\frac{1}{5.6.7}+...$$ = 0.166..+0.0166..+0.00476..+..... which is always <0.2 but a positive number.

Option A = $$e^2$$ - 1 = 7.34 - 1 = 6.34 therefore this cannot be the Answer

Option B = $$\log_{e}{2}-1$$ = $$\frac{log_{10}{2}}{log_{10}{e}}$$ - 1 = $$\frac{log_{10}{2}}{log_{10}{2.72}}$$ - 1

We know that the value of $$\frac{log_{10}{2}}{log_{10}{2.72}}$$ < 1 and therefore value of $$\frac{log_{10}{2}}{log_{10}{2.72}}$$ - 1 < 0

therefore this cannot be the Answer

Option C = $$2\log_{10}{2}-1$$ = $$\log_{10}{4}-1$$

We know that the value of $$\log_{10}{4}$$ < 1 and therefore value of $$\log_{10}{4}-1$$ < 0

therefore this cannot be the Answer

Therefore the Answer is option 'D' 


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