Question

If $P_m$ stands for $mP_m$, then $1+1p_1+2P_2+3P_3+…+n.P_n$ is equal to:

Answer 1

Here in the given question we are provided with the expression in which we need to obtain the summation of the provided series, here first we need to expand the series in the factorial form and then add the whole series, we know the properties of factorial here.

Complete step by step answer:
The given series first needed to be expressed in factorial form and then we can add the series, here the given series is of permutation, and we know to solve permutation we need to do factorial of the given term, on solving we get:
We have:

$$\Rightarrow {P_1} = {1P_1}=1! \\\ \Rightarrow {P_2} = {2P_2} = 2! \\\ \Rightarrow {P_3} = {3P_3} = 3! \\\ \Rightarrow {P_n} = {nP_n} = n! \\\$$

$nP_r= \dfrac{n!}{(n-r)!}$
Now we can write the expression as:

$$\Rightarrow 1 + 1{P_1} + 2{P_2} + 3{P_3} + ... + n{P_n} \\\ \Rightarrow 1 + (1)1! + (2)2! + (3)3! + ... + (n)n! \\\ \Rightarrow 1 + (2! - 1!) + (3! - 2!) + (4! - 3!) + ... + ((n + 1)! - n!) \\\ \Rightarrow (1 - 1!) + (2! - 2!) + (3! - 3!) + (4! - 4!) + ... + (n! - n!) + (n + 1)! \\\ \Rightarrow 0 + 0 + 0 + 0 + ... + (n + 1)! \\\ \Rightarrow (n + 1)! \\\$$

Here we express the term first and then rewrite the term so that the term remains the same value as it has earlier and accordingly adjust the values, here we see that after opening the brackets all the term cancel outs, except the last term which is the answer for the question.

Note: Here in the above question we rewrite the terms here one more method can be used which is, simply write all the terms and then see the series obtained and then put the summation sign on the general term of the series, and add for the required number of terms, the answer will remain same.