Question

The value of 2nP2+nPn-1 is:
​

Answer 1

$\sf {}^{2n} P_2 + {}^{n} P_{n - 1} = (4 {n}^{2} - 2n )+ n!$

Given : $\sf {}^{2n} P_2 + {}^{n} P_{n - 1}$

To find : The value

Solution :

Here the given expression is

$\sf {}^{2n} P_2 + {}^{n} P_{n - 1}$

We simplify it as below

$\sf {}^{2n} P_2 + {}^{n} P_{n - 1}$

$\displaystyle \sf{ = \frac{(2n)!}{(2n - 2)!} + \frac{n!}{(n - n + 1)!} }$

$\displaystyle \sf{ = \frac{2n(2n - 1)(2n - 2)!}{(2n - 2)!} + \frac{n!}{1!} }$

$\displaystyle \sf{ = 2n(2n - 1)+ n! }$

$\displaystyle \sf{ = (4 {n}^{2} - 2n )+ n! }$

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

$1.$ IF 1/1!9!+1/3!7!+1/5!5!+1/7!3!+1/9!=2^n /10! then n=?

https://brainly.in/question/20427754

2. If (1+x+x2)n = a0 +a1x +a2x2 + a3x3+ ...+ a2nx2n,show thata0+a3+a6 + .... = 3n-1

https://brainly.in/question/8768959