Prove 1(1!) + 2(2!)+3(3!)+...+n(n!)=(n+1)!-1 by using mathematical induction

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To prove,

1(1!) + 2(2!) + 3(3!) + ... + n(n!) = (n+1)! - 1

LHS = 1(1!) + 2(2!) + 3(3!) + ... + n(n!)

RHS = (n+1)! - 1

By mathematical induction:

**Let n = 1**,

Then, LHS = 1(1!) = 1 x 1 = 1

And RHS = (1 + 1)! - 1 = 2! - 1 = 2 - 1 = 1

So, both LHS and RHS = 1 and equation is true at n=1.

Now, second part is the assumption part where n=k holds true for the equation.

**Let n = k**,

The equation is assumed to be true, and is

1(1!) + 2(2!) + 3(3!) + ...... + k(k!) = (k+1)! - 1 ------------ (a)

Third part of induction is that, since the equation held true for k, it will hold true for k+1.

Let **n = k+1**,

RHS = [(k+1) + 1]! - 1

= (k+2)! - 1

LHS = 1(1!) + 2(2!) + 3(3!) + ...... + k(k!) + (k+1)[(k+1)!]

= {1(1!) + 2(2!) + 3(3!) + ...... + k(k!)} + (k+1)[(k+1)!]

= { (k+1)! - 1 } + (k+1)[(k+1)!] ------- substituting eq. (a).

= (k+1)! + (k+1)[(k+1)!] - 1 (simply re-written)

= (k+1)! [ 1 + (k+1) ] - 1

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

By formula, **a! x (a+1) = (a+1)!**

Therefore, LHS = (k+2)! - 1 and RHS = (k+2)! - 1

ie., LHS = RHS . The equation holds true for n=1, n=k and n=k+1

Hence the equation 1(1!) + 2(2!)+3(3!)+...+n(n!)=(n+1)!-1

is proved.

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