Question

summation from k=1 to 2 6k/k+13
write the sum without the sigma notation and evaluate it.
Help any1?

Answer 1

\[\sum_{k=1}^{2} \frac{ 6k }{ k+13 }\]

Answer 2

Sum from k=1 to 2.

So you only have two terms to add up.

At k=1: \(\dfrac{6k}{k+13}\to\dfrac{6(1)}{1+13} = \dfrac{6}{14}=\dfrac{3}{7}\)

Then for next and final term; k=2: \(\dfrac{6k}{k+13}\to\dfrac{6(2)}{2+13} = \dfrac{12}{15}=\dfrac{4}{5}\)

So you have sum \(\displaystyle\sum_{k=1}^2\dfrac{6k}{k+13}=\dfrac{3}{7}+\dfrac{4}{5}\) to evaluate.

Make senses so far?

Answer 3

thank you