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# Can anyone explain me this one?

#1
+9185
+3

The question is how to show that the above two products are equal to each other.

I don't know if this is the best way, but this is how I would do it:

\(\large {\prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k=1+m}^{k=n+m}a_k\)     because  \( \prod\limits_{k=1+m}^{n+m}a_k\)   means the same as  \( \prod\limits_{k=1+m}^{k=n+m}a_k\)

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k-m=1}^{k-m=n}a_k\)      by subtracting  m  from both sides of each equation

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k-m=1}^{k-m=n}a_{(k-m+m)}\)         since  k - m + m  =  k

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{{\color{RubineRed}k-m}=1}^{{\color{RubineRed}k-m}=n}a_{({\color{RubineRed}k-m}+m)}\)

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k=1}^{k=n}a_{(k+m)}\)       because we can replace the pink text with whatever we want

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k=1}^{n}a_{(k+m)}\)

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Dec 24, 2020

#1
+9185
+3

The question is how to show that the above two products are equal to each other.

I don't know if this is the best way, but this is how I would do it:

\(\large {\prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k=1+m}^{k=n+m}a_k\)     because  \( \prod\limits_{k=1+m}^{n+m}a_k\)   means the same as  \( \prod\limits_{k=1+m}^{k=n+m}a_k\)

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k-m=1}^{k-m=n}a_k\)      by subtracting  m  from both sides of each equation

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k-m=1}^{k-m=n}a_{(k-m+m)}\)         since  k - m + m  =  k

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{{\color{RubineRed}k-m}=1}^{{\color{RubineRed}k-m}=n}a_{({\color{RubineRed}k-m}+m)}\)

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k=1}^{k=n}a_{(k+m)}\)       because we can replace the pink text with whatever we want

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k=1}^{n}a_{(k+m)}\)

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hectictar Dec 24, 2020