gorgeousgen9487

2022-07-02

Can $\frac{{b}_{0}}{{a}_{0}}+\frac{{b}_{1}}{{a}_{1}}+\frac{{b}_{2}}{{a}_{2}}+\frac{{b}_{3}}{{a}_{3}}+...+\frac{{b}_{n}}{{a}_{n}}$ be represented as ...
Is this correct? (Last step $\to$ After taking L.C.M.)
$\frac{{b}_{0}}{{a}_{0}}+\frac{{b}_{1}}{{a}_{1}}+\frac{{b}_{2}}{{a}_{2}}+\frac{{b}_{3}}{{a}_{3}}+...+\frac{{b}_{n}}{{a}_{n}}=\sum _{k=0}^{n}\left(\frac{{b}_{k}}{{a}_{k}}\right)=\frac{\sum _{p=0}^{n}\left(\frac{{b}_{p}}{{a}_{p}}×\prod _{q=0}^{n}{a}_{q}\right)}{\prod _{q=0}^{n}{a}_{q}}$
Thanks!

Brendan Bush

Expert

This is true simply by linearity of summation: if $k$ is constant with respect to $p$ then
$\sum _{p=0}^{n}k{a}_{p}=k\cdot \sum _{p=0}^{n}{a}_{p}$
Here, $k=\prod _{q=0}^{n}{a}_{q}$, which is constant with respect to $p$, so you can factor it out of the sum and cancel it.
You mention LCMs, though it doesn't look like you used them anywhere. In any case, your equation remains true if you replace $\prod _{q=0}^{n}{a}_{q}$ by $\mathrm{l}\mathrm{c}\mathrm{m}\left\{{a}_{q}\mid 1\le q\le n\right\}$

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