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$$\log_{10}{10} + \log_{10}{10^2} + ..... + \log_{10}{10^n}$$
SinceΒ $$\log_aa\ $$ = 1
$$\log_{10}{10} + \log_{10}{10^2} + ..... + \log_{10}{10^n}$$ = 1+2+....n
=$$\ \frac{\ n\left(n+1\right)}{2}$$Β
=$$\frac{(n^{2} + n)}{2}$$
D is the correct answer.
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