How to Evaluate This Interesting Expression?

[anemone]

Evaluate $1\cdot 2^2 + 1\cdot 2\cdot 3^2 + 1\cdot 2\cdot 3\cdot 4^2 +\cdots+ 1\cdot 2\cdot 3\cdots 2015^2− (1\cdot 2\cdot 3\cdots 2016)$.

 

[MarkFL]

My solution:

Spoiler

First, let's prove by induction the following hypothesis $P_n$:

$$\sum_{k=a}^n(k\cdot k!)=(n+1)!-a!$$

1.) The base case $P_a$:

$$\sum_{k=a}^n(k\cdot k!)=a\cdot a!=a!((a+1)-1)=(a+1)!-a!$$

The base case is true.

2.) The induction step:

$$\sum_{k=a}^n(k\cdot k!)=(n+1)!-a!$$

Add through by $$(n+1)(n+1)!$$:

$$\sum_{k=a}^n(k\cdot k!)+(n+1)(n+1)!=(n+1)!-a!+(n+1)(n+1)!$$

$$\sum_{k=a}^{n+1}(k\cdot k!)=(n+1)!((n+1)+1)-a!$$

$$\sum_{k=a}^{n+1}(k\cdot k!)=((n+1)+1)!-a!$$

We have derived $P_{n+1}$ from $P_n$ thereby completing the proof by induction.

And so we may now state:

$$S=\sum_{k=2}^{2015}(k\cdot k!)-2016!=(2015+1)!-2!-2016!=-2$$

 

[anemone]

My solution:

Spoiler

First, let's prove by induction the following hypothesis $P_n$:

$$\sum_{k=a}^n(k\cdot k!)=(n+1)!-a!$$

1.) The base case $P_a$:

$$\sum_{k=a}^n(k\cdot k!)=a\cdot a!=a!((a+1)-1)=(a+1)!-a!$$

The base case is true.

2.) The induction step:

$$\sum_{k=a}^n(k\cdot k!)=(n+1)!-a!$$

Add through by $$(n+1)(n+1)!$$:

$$\sum_{k=a}^n(k\cdot k!)+(n+1)(n+1)!=(n+1)!-a!+(n+1)(n+1)!$$

$$\sum_{k=a}^{n+1}(k\cdot k!)=(n+1)!((n+1)+1)-a!$$

$$\sum_{k=a}^{n+1}(k\cdot k!)=((n+1)+1)!-a!$$

We have derived $P_{n+1}$ from $P_n$ thereby completing the proof by induction.

And so we may now state:

$$S=\sum_{k=2}^{2015}(k\cdot k!)-2016!=(2015+1)!-2!-2016!=-2$$

Very well done MarkFL!(Cool)

 

[kaliprasad]

Spoiler

1st let us evaluate the sum

1st we see that $n^{th}$ term = $n * n! = (n+1-1) * n!= (n+1)! - n!$

when we sum the above from 2 to 2015 we get as a telescopic sum 2016!- 2 !
subtracting the last value that is 2016! we are left with -2! or - 2

 

[anemone]

Thanks for participating, kaliprasad!

For your information, that is exactly how I approached this particular problem as well! (Cool)