यदि`P_m = ""^(m)P_(m)` तो सिद्ध कीजिए कि `1+1.P_(1)+2.P_(2)+3.P_(3)+"....."+n.P_(n)=(n+1)!`

Explanation

प्रश्नानुसार,`P_(m) = ""^(m)P_(m)=m!`
इसलिए `1+1.P_(1) +2.P_(2)+3.P_(3)+"...."+n.P_(n)= 1+ 1! +2.2! +3.3! +"..."+n.n!`
`=1+underset(r=1)overset(n)sumr.r! =1+underset(r=1)overset(n)sum[(r+1)-1]r!`
`= 1+underset(r=1)overset(n)sum[(r+1)r!-r!]`
`rArr 1+underset(r=1)overset(n)sum[(r+1)!-r!]`
`=1+[(2!-1!)+(3!-2!)+(4!-3!)+"..."+{(n+1)!-n!}]`
`= 1 + [(n+1)!-1!]=(n+1)!`