Question

the value of sigma r=1 to 10 r×rpr​

Answer 1

Answer:

P

n

n

=

(n−1)!

n!

=n!

So, ∑

r=1

10

rP

r

r

=∑

r=1

10

r(r!)=∑

r=1

10

(r+1)r!−r!

∑

r=1

10

rP

r

r

=2×1!−1!+3×2!−2!+4×3!−3!+........+11×10!−10!

So, ∑

r=1

10

rP

r

r

=11×10!−1!=11!−1

Step-by-step explanation:

I hope it will help you

Answer 2

${P}^{n}_n=\frac{n!}{(n-1)!}=n!\\\\so,{\sum}^{10}_{r=1} r{P}^{r}_r ={ \sum}^{10}_{r=1}r(r!)\\\\={ \sum}^{10}_{r=1}(r+1)r!-r!\\\\{ \sum}^{10}_{r=1}r{P}^{r}_r=2×1!-1!+3×2!-2!+\\4×3!-3!+....+11×10!-10!\\\\so,{\sum}^{10}_{r=1}r{P}^{r}_r=11×10!-1!=11!-1!$