Ask Question
26 December, 08:13

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

+4
Answers (1)
  1. 26 December, 08:15
    0
    Step-by-step explanation:

    Let's assume that

    P (n) = 1.1! + 2.2! + 3.3! + ... + n. n! = (n + 1) ! - 1.

    For n = 1

    L. H. S = 1.1!

    = 1

    R. H. S = (n + 1) ! - 1.

    = (1 + 1) ! - 1.

    = 1

    L. H. S = R. H. S

    Hence the P (n) is true for n=1

    Fort n = 2

    L. H. S=1.1! + 2.2!

    =1+4

    =5

    R. H. S = (2 + 1) ! - 1.

    = (2 + 1) ! - 1.

    = 5

    L. H. S = R. H. S

    Hence the P (n) is true for n=2

    Let'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]!-1

    So, P (n) is also true for (n+1).

    So, P (n) is true for all integers n.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Use induction to prove that for all integers n 2 1 we have 1.1! + 2.2! + 3.3! + ... + nin! = (n + 1) ! - 1. ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers