Lawrence's sum

Let $P(n) = \dfrac{n+2}{n! + (n+1)! + (n+2)!}$ . We can use some algebra to rework this expression: $\dfrac{n+2}{n! + (n+1)! + (n+2)!} = \dfrac{n+2}{n! + n!(n+1) + n!(n+1)(n+2)} =$

$= \dfrac{n+2}{n!(1 + (n+1) + (n+1)(n+2))} = \dfrac{n+2}{n!((n+2) + (n+1)(n+2))} =$

$= \dfrac{n+2}{n!(n+2)^2} = \dfrac 1{n!(n+2)} = \dfrac{n+1}{(n+2)!} = \dfrac{n+2}{(n+2)!} - \dfrac1{(n+2)!} =$

$= \dfrac1{(n+1)!} - \dfrac1{(n+2)!}$

When the sum on the question is extended, we have: $P(1) + P(2) + P(3) + \dots + P(2111) =$ $= \dfrac12 - \dfrac1{3!} + \dfrac1{3!} - \dfrac1{4!} + \dfrac1{4!} - \dots - \dfrac1{2112!} + \dfrac1{2112!} - \dfrac1{2113!}$

Using the telescoping sum technique, we have: $\frac12 - \frac1{2113!}$ . Then, $a = 2113$ and the answer is $113$ .

In a problem like this, it really helps if the final answer form is given. In this case, it is $\frac{1}{2}-\frac{1}{a!}$ . This means that this is a telescopic series. So, we have to simplify the given expression as a difference of two consecutive terms. This was the motivation for my approach.

We have $\frac{n+2}{n!+(n+1)!+(n+2)!}$ . Consider the denominator. We will simply it first.

$n!+(n+1)!+(n+2)!= n!\times [1+(n+1)+(n+1)(n+2)]=n!(1+n+1+n^2+3n+2)$

$\Rightarrow n!+(n+1)!+(n+2)!=n!\times (1+n+1+n^2+3n+2)=n!\times (n^2+4n+4)=n!\times(n+2)^2$

Hence, our expression $\frac{n+2}{n!+(n+1)!+(n+2)!}=\frac{n+2}{n!\times(n+2)^2}=\frac{1}{n!\times(n+2)}$

$\Rightarrow \frac{1}{n!\times(n+2)}=\frac {n+1}{n!(n+1)(n+2)}=\frac{n+1}{(n+2)!}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}$

So we have $\displaystyle \sum_{n=1}^{2111}\frac{n+2}{n!+(n+1)!+(n+2)!}=\displaystyle \sum_{n=1}^{2111}[\frac{1}{(n+1)!}-\frac{1}{(n+2)!}]$

Plugging in values of $n$ gives us

$\displaystyle \sum_{n=1}^{2111}[\frac{1}{(n+1)!}-\frac{1}{(n+2)!}]=(\frac{1}{2!}-\frac{1}{3!})+(\frac{1}{3!}-\frac{1}{4!})...+(\frac{1}{2112!}-\frac{1}{2113!})$

As is evident, all consecutive terms (other than the first and the last terms) get cancelled out.

So $\displaystyle \sum_{n=1}^{2111}\frac{n+2}{n!+(n+1)!+(n+2)!}=\frac{1}{2!}-\frac{1}{2113!}=\frac{1}{2!}-\frac{1}{a!}\Rightarrow a=2113$

So the answer is $\boxed{113}$

First, we would try to simplify the $k^{th}$ term of the sequence so that we break it into the difference of two terms, such that we could easily sum of the series.

Now take $k!$ common from the denominator, we have -

$\frac{k+2}{k!}\times\frac{1}{1+(k+1)+(k+1)(k+2)}$

$=\frac{k+2}{k!}\times\frac{1}{(k+2)^2}$

$=\frac{1}{k!\times(k+2)}$

Now multiplying the numerator and denominator by $k+1$ gives us - $\frac{k+1}{(k+2)!}$

which can be written as - $\frac{k+2}{(k+2)!}-\frac{1}{(k+2)!} \Rightarrow \frac{1}{(k+1)!}-\frac{1}{(k+2)!}$

Now we just need to sum the series -

$\sum_{k=1}^{2111} \frac{1}{(k+1)!}-\frac{1}{(k+2)!}$

$=\bigg(\frac{1}{2!}-\frac{1}{3!}\bigg)+\bigg(\frac{1}{3!}-\frac{1}{4!}\bigg)+ \ldots +\bigg(\frac{1}{2112!}-\frac{1}{2113!}\bigg)$

$=\frac{1}{2!}-\frac{1}{2113!}$

Hence $a=2113$ . So the last three digits of $a$ would be $\boxed{113}$

The approach is to try small cases. If we let the upper limit of summation be $k$ , then we have $k = 1 = \dfrac{1}{3} = \dfrac{1}{2} - \dfrac{1}{3!}$ $k = 2 = \dfrac{1}{3} + \dfrac{1}{8} = \dfrac{11}{24} = \dfrac{1}{2} - \dfrac{1}{4!}$ $k = 3 = \dfrac{1}{3} + \dfrac{1}{8} + \dfrac{1}{30} = \dfrac{59}{120} = \dfrac{1}{2} - \dfrac{1}{5!}$ Clearly, there is a pattern: $a = k+2$ . We can prove this by induction. Thus, our answer is $2111 + 2 = 2113 \Rightarrow \boxed{113}$ .

Checking the denominator:

$n! + (n+1)! + (n+2)! = n!(1 + n+1 + (n+!)(n+2)$

= $n!(n+2)(1 + n+ 1) = n!(n+2)^2$

Now , $\frac{n+2}{n!(n+2)^2} = \frac{1}{n!(n+2)}$

Multiply and divide by $n+1$ to get

$\frac{n+1}{(n+2)!} = \frac{(n+2) - 1}{(n+2)!} = \frac{1}{(n+1)!} - \frac{1}{(n+2)!}$

Clearly, applying summation, the sum telescopes to $\frac{1}{2!} - \frac{1}{2113!}$

Hence the answer is $2\fbox{113}$

$n!+(n+1)!+(n+2)!=n!\big(1+(n+1)+(n+1)(n+2)=n!(n+2)^2$ .

So $\displaystyle\frac{n+2}{n!+(n+1)!+(n+2)!}=\frac{1}{n!(n+2)}=\frac{n+1}{(n+2)!}=\frac{(n+2)-1}{(n+2)!}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}$ . This series telescopes.

Now $\sum_{n=1}^{2111} \displaystyle\frac{1}{(n+1)!}-\frac{1}{(n+2)!}=\frac{1}{2!}-\frac{1}{2113!}=\frac{1}{2!}-\frac{1}{a!}$ .

So comparing last three digits are $\boxed{113}$ .

Using the fact that $n! = n(n-1)!$ , we can write the sum as :

$S = \displaystyle \sum_{n=1}^{2111} \frac{n+2}{n! + (n+1)n! + (n+2)(n+1)n!}\\ = \displaystyle \sum_{n=1}^{2111} \frac{n+2}{n!(1 + (n+1) + (n+1)(n+2))} \\= \displaystyle \sum_{n=1}^{2111} \frac{n+2}{n!((n + 2) + (n+1)(n+2))} = \displaystyle \sum_{n=1}^{2111} \frac{n+2}{n!(n + 2)(1 + n+1)} \\ =\displaystyle \sum_{n=1}^{2111} \frac{1}{n!(n+2)}$

Using some algebraic manipulations , we can write :

$S = \displaystyle \sum_{n=1}^{2111} \frac{1}{n!(n+2)} = \displaystyle \sum_{n=1}^{2111} \frac{n+1}{n!(n+1)(n+2)} = \displaystyle \sum_{n=1}^{2111} \frac{n+1}{(n+2)!} \\ = \displaystyle \sum_{n=1}^{2111} \frac{n +1+1-1}{(n+2)!} = \displaystyle \sum_{n=1}^{2111} \frac{n + 2-1}{(n+2)!} = \displaystyle \sum_{n=1}^{2111} \left( \frac{n+2}{(n+2)!} - \frac{1}{(n+2)!} \right) \\ \therefore S = \displaystyle \sum_{n=1}^{2111} \left( \frac{1}{(n+1)!} - \frac{1}{(n+2)!} \right)$

The sum is now in the form of a telescoping sum , therefore :

$S = \frac{1}{(1+1)!} - \frac{1}{((2111 + 1) +1)!} = \frac{1}{2} - \frac{1}{(2113)!} \\ ,\because S = \frac{1}{2} - \frac{1}{a!} \\ \therefore a = 2113$

$\therefore$ The last three digits of $a$ are $113$

What I did is that I just checked The pattern I put 1 in the above equation than checked the value of a. a=3. than for n= 2 in the above equation I check the value of a , it came out to be 4, so for n = n the value of a should be a=n+2 so for n= 2111 the value of a should be a = 2113

We want to compute Sum 1/(n! (n+2)) let f(i) = 1/[(n+2)!] so -(f(i)-f(i-1)) = 1/(n! (n+2)) then it is easy to use a telescopic

FROM THE DENOMINATOR TAKE n! COMMON AND FACTORISE IT TO GET $(n+2)^{2}$

ONE OF THE (n+2) WILL BE CANCELLED.

MULTIPLY (n+1) IN NUMERATOR AND DENOMINATOR

DENOMINATOR CHANGES TO (n+2)!

CHANGE n+1=n+2-1

FACTORISE TO GET 1/(n+1)!-1/(n+2)!

PUT SOME VALUES TO OBTAIN A TELESCOPING SERIES

ANSWER-1/2!-1/2113!

SO $\boxed{113}$

Taking n=1 sum = (1/3) = (1/2) - (1/6)... therefore a! = 3! or a=3

Taking n=2 sum = (11/24) = (1/2) - (1/24)... therefore a! = 4! or a=4

Taking n=3 sum = (59/120) = (1/2) - (1/120)... therefore a! = 5! or a=5

As seen above, the diff between n and a is 2 Therfore for n=2111, a will be 2111+2=2113 Therefore the entered answer has to be 113

[(n+2)/n!+(n+1)!+(n+2)!]=1/n!(n+2) [on solving] 1/n!(n+2)=1/[n!+(n+1)!]=1/(n+1)!-1/(n+2)! now sum(1 to 2111)[(n+2)/n!+(n+1)!+(n+2)!]=(1/2!-1/3!)+(1/3!-1/4!)+........................(1/2112!-1/2113!) =1/2-1/2113! so last three digit of 2113 is 113