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If you type:

2 n1 p[n1-1, n2] /. {n1 - 1 -> z1}

then the answer is:

2 n1 p[z1, n2]

This is shorthand for the function:

ReplaceAll[2 n1 p[n1 - 1, n2], n1 - 1 -> z1]

which explicitly uses the rule (in the second argument) to change the first argumemnt.

If you type:

2 n1 p[n1-1, n2] /. {n1 -> z1 +1}

then the answer is:

2 (1 + z1) p[z1, n2]

This is shorthand for the function:

ReplaceAll[2 n1 p[n1 - 1, n2], n1 -> z1+1]

which explicitly uses the rule (in the second argument) to change the first argumemnt.

If you type:

2 ∗n1∗p[n1-1, n2] /. {n1 - 1 -> z1}

then the answer is:

2 ∗n1∗p[z1, n2]

This is shorthand for the function:

ReplaceAll[2 ∗n1∗p[n1 - 1, n2], n1 - 1 -> z1]

which explicitly uses the rule (in the second argument) to change the first argumemnt.

If you type:

2 n1 p[n1-1, n2] /. {n1 - 1 -> z1}

then the answer is:

2 n1 p[z1, n2]

This is shorthand for the function:

ReplaceAll[2 n1 p[n1 - 1, n2], n1 - 1 -> z1]

which explicitly uses the rule (in the second argument) to change the first argumemnt.

If you type:

2 ∗n1∗p[n1-1, n2] /. {n1 - 1 -> z1}

then the answer is:

2 ∗n1∗p[z1, n2]

is this not what you want?

If you type:

2 ∗n1∗p[n1-1, n2] /. {n1 - 1 -> z1}

then the answer is:

2 ∗n1∗p[z1, n2]

This is shorthand for the function:

ReplaceAll[2 ∗n1∗p[n1 - 1, n2], n1 - 1 -> z1]

which explicitly uses the rule (in the second argument) to change the first argumemnt.