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Question: Let \[{{P}_{m}}\] stand for \[^{n}{{P}_{m}}\]. Then the expression \[1\cdot {{P}_{1}}+2\cdot {{P}_{2...

Let Pm{{P}_{m}} stand for nPm^{n}{{P}_{m}}. Then the expression 1⋅P1+2⋅P2+3⋅P3+....+n⋅Pn1\cdot {{P}_{1}}+2\cdot {{P}_{2}}+3\cdot {{P}_{3}}+....+n\cdot {{P}_{n}} is equal to
A.(n+1)!−1\left( n+1 \right)!-1
B. (n+1)!+1\left( n+1 \right)!+1
C. (n+1)!\left( n+1 \right)!
D. None of these.

Explanation

Solution

In this problem, we have to find the value of the given expression. We can first write the given expression. We can then write them in the relevant factorial form. We can then convert the step into summation format and simplify it step by step and we can simplify the factorial format to get the value of the given expression.

Complete step by step answer:
Here we have to find the value of the given expression.
We know that the expression given is,
1⋅P1+2⋅P2+3⋅P3+....+n⋅Pn1\cdot {{P}_{1}}+2\cdot {{P}_{2}}+3\cdot {{P}_{3}}+....+n\cdot {{P}_{n}}
We are also given Pm{{P}_{m}} and we can write it as,
Pm=nPm=m!{{P}_{m}}{{=}^{n}}{{P}_{m}}=m! ……. 1)
From (1), we can now write the given expression as,
⇒1⋅P1+2⋅P2+3⋅P3+....+n⋅Pn=1⋅1!+2⋅2!+3⋅3!+....+n⋅n!\Rightarrow 1\cdot {{P}_{1}}+2\cdot {{P}_{2}}+3\cdot {{P}_{3}}+....+n\cdot {{P}_{n}}=1\cdot 1!+2\cdot 2!+3\cdot 3!+....+n\cdot n!
We can now rewrite the above step in summation format, we get
⇒∑r=1nr⋅r!=∑r=1n[(r+1)−1]⋅r!\Rightarrow \sum\limits_{r=1}^{n}{r\cdot r!}=\sum\limits_{r=1}^{n}{\left[ \left( r+1 \right)-1 \right]\cdot r!}
We can now write the above step as,
⇒∑r=1nr⋅r!=∑r=1n[(r+1)⋅r!−r!]=∑r=1n[(r+1)!−r!]\Rightarrow \sum\limits_{r=1}^{n}{r\cdot r!}=\sum\limits_{r=1}^{n}{\left[ \left( r+1 \right)\cdot r!-r! \right]}=\sum\limits_{r=1}^{n}{\left[ \left( r+1 \right)!-r! \right]}
We can now substitute the value of r like given in the summation in the above step, we get
⇒[(2!−1!)+(3!−2!)+(4!−3!)+.......+(n+1)!−n!]\Rightarrow \left[ \left( 2!-1! \right)+\left( 3!-2! \right)+\left( 4!-3! \right)+.......+\left( n+1 \right)!-n! \right]
We can now simplify the above step, we get
⇒(n+1)!−1!\Rightarrow \left( n+1 \right)!-1!
The value of the expression 1⋅P1+2⋅P2+3⋅P3+....+n⋅Pn=(n+1)!−1!1\cdot {{P}_{1}}+2\cdot {{P}_{2}}+3\cdot {{P}_{3}}+....+n\cdot {{P}_{n}}=\left( n+1 \right)!-1!

So, the correct answer is “Option A”.

Note: We should always remember that factorial is a function that multiplies a number by every number below it. We should also know how to write the summation format as we write which value to be substituted in the below and up to which values can be substituted in the above.