Discussion

Explanation:

Let S = $\frac{1}{2}$ + $\frac{1}{2+4}$ + $\frac{1}{2+4+6}$ + ...

Tn = $\frac{1}{2+4+...+2\mathrm{n}}$ = $\frac{1}{2(1+2+...+\mathrm{n})}$ = $\frac{1}{\mathrm{n}(\mathrm{n}+1)}$ = $\frac{1}{\mathrm{n}}$ - $\frac{1}{\mathrm{n}+1}$

⇒ S = $\left(\frac{1}{1}-\frac{1}{2}\right)$ + $\left(\frac{1}{2}-\frac{1}{3}\right)$ + $\left(\frac{1}{3}-\frac{1}{4}\right)$ + ... + $\left(\frac{1}{9}-\frac{1}{10}\right)$ + $\left(\frac{1}{10}-\frac{1}{11}\right)$

⇒ S = 1 - $\frac{1}{11}$ = $\frac{10}{11}$

Hence, option (c).

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