Evaluate Logarithmic Sum

Q.11. Evaluate

∑n=1100log10⁡(31−n)

Answer Choices

Solution

We use the logarithmic property logb⁡(ac)=clogb⁡(a) to simplify the sum:

log10⁡(31−n)=(1−n)log10⁡(3)

This changes the sum to:

∑n=1100(1−n)log10⁡(3)=log10⁡(3)∑n=1100(1−n)

We split the sum:

∑n=1100(1−n)=∑n=11001−∑n=1100n=100−5050=−4950

Thus, the sum becomes:

−4950log10⁡(3)

The correct answer is: −4950log10⁡(3) .