**Let ππ(ππ) be the proposition that 1(1!) + 2(2!) + 3(3!) + β―+ ππ(ππ!) = (ππ+ 1)! β1. Prove by induction that ππ(ππ) is true for all ππβ₯1.**

The **Answer to the Question**

is below this banner.

**Here's the Solution to this Question**

LetΒ $P(n)$Β be the proposition that

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

BASIS STEP:

$P(1)$Β is true, becauseΒ $1(1!)=1=(1+1)!-1.$

INDUCTIVE STEP:

For the inductive hypothesis we assume thatΒ $P(k)$Β holds for an arbitrary

positive integerΒ $k.$Β That is, we assume that

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

Under this assumption, it must be shown thatΒ $P(k + 1)$Β is true, namely, that

$1(1!)+2(2!)+3(3!)+...+k(k!)+(k+1)(k+1)!$

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

When we addΒ $(k+1)(k+1)!$Β to both sides of the equation inΒ $P(k),$Β we obtain

$1(1!)+2(2!)+3(3!)+...+k(k!)+(k+1)(k+1)!$

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

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

This last equation shows thatΒ $P(k + 1)$Β is true under the assumption thatΒ $P(k)$Β is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we know thatΒ $P(n)$Β is true for all positive integersΒ $n.$

That is, we have proven that

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

for all positive integersΒ $n.$