A factorial sum

• $\displaystyle\sum_{i=1}^{2014}n\times n!$
• $\displaystyle=\sum_{i=1}^{2014}[n\times n!+n!-n!]$
• $\displaystyle=\sum_{i=1}^{2014}[(n+1)\times n!-n!]$
• $\displaystyle=\sum_{i=1}^{2014}[(n+1)!-n!]$
• $\displaystyle=\sum_{i=1}^{2014}(n+1)!-\sum_{i=1}^{2014}n!$
• $\displaystyle=\sum_{i=2}^{2015}n!-\sum_{i=1}^{2014}n!$
• $\displaystyle=2015!-1!$
• $\displaystyle=2015!-1$

Therefore $n=\fbox{2015}$ .

More about telescoping series can be found here .

The problem can be written as: Summation of n*(n!) from 1 to 2014

Now, n (n!) = {(1+n-1)/(n+1)} {(n+1)*(n!)}

=[1-{1/(n+1)}]*(n+1)!

=(n+1)! - (n)!

So the expression can be written:

2! - 1! + 3! - 2! + 4! - 3! + 5! - 4! .............. + 2015! - 2014!

=2015! - 1!

So, n = 2015

Now, seemingly complicated series are actually the simplest. Whenever, you find Sigma/Summation associated with huge numbers, sums or, products, try converting the expression into a Difference of two consecutive terms. ( sometimes, Sum of Differences of Consecutive terms ) The summation thereby, gets reduced to the difference of The Last Term & The First Term. Mathematically, Sigma [ Tn+1 - Tn ] = T(Upper Limit) - T(Lower Limit)

Here. n.n! = (n+1-1) n! = (n+1) n! - n! = (n+1)! - n! So, Sum = (2014+1)! - 1! = 2015! - 1..........................:)

I shall do this by induction.... Let us assume that there is only one term on LHS I.e. 1×1!=1....then to make LHS and RHS equal, n should be qual to 2. Now what we have seen is that as there are number of terms in LHS, the value of n=number of terms in LHS+1. Now let us assume the equation is true for number of terms in LHS=k.... Then, 1×1!+2×2!+3×3!.........k×k!=(k+1)! -1....... Now we shall prove this equation to be true for k+1 number of terms.....So let us add (k+1)[(k+1)!] to both sides, thus the equation becomes 1×1!+2×2!+3×3!.........k×k!+(k+1)[(k+1)!]=(k+1)! -1+(k+1)[(k+1)!]=(k+1)!+(k+1)[(k+1)!]-1= (k+1)! -1=(k+2)!-1..... Hence we have seen that the equation holds true.....so for 2014 terms n=2015

answer is n+1 if the last term of series is n(n!)

n.n!=(n+1-1)n!=(n+1)n!-n!=(n+1)!-n! So, 1.1!=2!-1! 2.2!=3!-2! .... ...... 2013.2013!=2014!-2013!

2014.2014!=2015!-2014!

Summing all, it's 2015!-1

According to me, it doesn't feel that lame if you just pattern match the problem given to you, just a coincidence as it works, but never pattern match and answer, since you have to solve to manually work out all the cases...

when you take the 1, or say 1! from the rhs to lhs; eqn becomes: 2.1!+2.2!+3.3!+...+2014.(2014)!=n! taking 2 common out of first two terms, our expression becomes: 2(1!+2!)+3.3!+4.4!+.... => (3.2!+3.3!+4.4!+..... => (3.(2!+3!)+4.4!+..... => 4.3!+4.4!+5.5!+..... by a similar logic: .....2013.(2012)!+2013.(2013)!+2014.(2014)!=n! => 2014.(2013)!+2014.(2014)!=n! => 2014(2013!+2014!)=n! => n!=2015.(2014)! => n!=2015! => n=2015