MATH SOLVE

4 months ago

Q:
# Use induction to prove that for all integers n 2 1 we have 1.1! +2.2! +3.3! + ... +nin! = (n + 1)! - 1.

Accepted Solution

A:

Step-by-step explanation:Let's assume thatP(n)=1.1! +2.2! +3.3! + ... +n.n! = (n + 1)! - 1.For n = 1L.H.S = 1.1! = 1R.H.S = (n + 1)! - 1. =(1 + 1)! - 1. = 1L.H.S = R.H.SHence the P(n) is true for n=1Fort n = 2L.H.S=1.1! +2.2! =1+4 =5R.H.S = (2 + 1)! - 1. =(2 + 1)! - 1. = 5L.H.S = R.H.SHence the P(n) is true for n=2Let's assume that P(n) is true for all n.Then we have to prove that P(n) is true for (n+1) too.So,L.H.S = 1.1! +2.2! +3.3! + ... +n.n!+(n+1).(n+1)! = (n + 1)! - 1 +(n+1).(n+1)! = (n+1)![1+(n+1)]-1 =(n+1)!(n+2)-1 =(n+2)!-1 =[(n+1)+1]!-1So, P(n) is also true for (n+1).So, P(n) is true for all integers n.