Question

Prove by the principle of mathematical induction that 1*1! + 2*2! +3*3! + ......... + n*n! = ( n +1 ) ! - 1 for all natural numbers n

Answer 1

Answer:

Induction method involves two steps, One, that the statement is true for n =1 ... First Step − Now for 11⋅2+12⋅3+. ... Hence, given statement is true for n=1 and n= 2 .

Answer 2

Answer:

STEP 1: Let P(n) be the given statement.

i.e.P(n): 1 × 1 ! + 2 × 2 ! + 3 × 3 !

+ .... + n × n ! = (n + 1)! - 1.

STEP 2: For n = 1,we have

L.H.S = 1 × 1 ! = 1

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

∵ L.H.S = R.H.S

∴ P(1) is true.

STEP 3: Let us assume that P(n) is true for n = K.

Then, we have

P(k) : 1 × 1 ! + 2 × 2 ! + 3 × 3 ! + ... + k × k !

= (k + 1) ! - 1 ... ---> 1

STEP 4: Now, we shall prove to show that

n = k + 1 .

For this we have to show that

1 × 1 ! + 2 × 2 ! + 3 × 3 ! + ... + k × k !

+ (k + 1) × (k + 1) ! = (k + 1 + 1) ! - 1

Then, L.H.S = 1 × 1 ! + 2 × 2 ! + 3 × 3 + ... + k × k !

+ (k + 1) × (k + 1) !

= (k + 1) ! - 1 + (k + 1) ! × (k + 1) [ From Eq. 1 ]

= (k + 1 + 1) (k + 1) ! - 1 = (k + 2) (k + 1) ! - 1

= (k + 2) ! - 1 = (k + 1 + 1) ! [ ∵ n(n - 1)! = n ! ]

Thus, P(k + 1) is true, whenever P(k) is true.

Hence, by the principal of mathematical

induction,

P(n) is true for all natural number n.

Step-by-step explanation:

@SSR