Answer to Question #273408 in Discrete Mathematics for haji

Question

Let 𝑃𝑃(𝑛𝑛) be the proposition that 1(1!) + 2(2!) + 3(3!) + ⋯+ 𝑛𝑛(𝑛𝑛!) = (𝑛𝑛+ 1)! −1. Prove by induction that 𝑃𝑃(𝑛𝑛) is true for all 𝑛𝑛≥1.

Answer 1

Let $P(n)$ be the proposition that

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

BASIS STEP:

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

INDUCTIVE STEP:

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

positive integer $k.$ That is, we assume that

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

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

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

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

When we add $(k+1)(k+1)!$ to both sides of the equation in $P(k),$ 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 that $P(k + 1)$ is true under the assumption that $P(k)$ is true. This completes the inductive step.

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

That is, we have proven that

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

for all positive integers $n.$