Trailing zeroes of a sum

$n= 1 + \displaystyle \sum_{k=1}^{2013} k!( (k+1)^3 - k^2)$

= $1 + \displaystyle \sum_{k=1}^{2013} \big( (k+1)! (k+1)^2 - k!k^2 \big)$

= $1 + (2014)^2 (2014)! - 1$ = $\boxed{(2014)^2 2014!}$

Now, no. of trailing zeroes = exponent of 5 = $\displaystyle \sum_{r=1}^{4} \bigg \lfloor \frac{2014}{5^r} \bigg \rfloor = \fbox{501}$

Calculating the sum $\sum_{k=1}^{2013}k!(k^3+2k^2+3k+1)$ in a direct manner would not be an easy task. So, would like to telescope the series, so we can easily find the sum

Now, observe that $k^3+2k^2+3k+1 = (k+1)^3-k^2$

Hence, $\sum_{k=1}^{2013}k!(k^3+2k^2+3k+1) = \sum_{k=1}^{2013}k!((k+1)^3-k^2)$

$= \sum_{k=1}^{2013}(k+1)^3k!-k^2k! = \sum_{k=1}^{2013}(k+1)^2(k+1)!-k^2k!$

Now it's easy to calculate the sum, $= (2^22!-1^21!)+(3^23!-2^22!)+\ldots(2014^2-2014!-2013^22013!) = 2014^22014!-1^2 1!$

Thus, $n=1+2014^22014!-1=2014^2 2014!$

We know, that there will be no trailing zeroes in the decimal representation of $2014^2$ . And so we now need to know the trailing zeroes in $2014!$ , which is similar to calculating exponent of $10$ (or $5\times2$ ) in $2014!$

Since, exponent of $5$ will be less than that of $2$ , so it's enough to calculate the exponent of $5$ , which is

$=[\frac{2014}{5}]+[\frac{2014}{5^2}]+[\frac{2014}{5^3}]+[\frac{2014}{5^4}]+\ldots=501$

Thus there will be $\boxed{501}$ trailing zeroes in decimal representation of $n=1+\sum_{k=1}^{2013}k!(k^3+2k^2+3k+1)$

$\displaystyle \sum_{k=1}^{2013} k!(k^3+2k^2+3k+1) = \displaystyle \sum_{k=1}^{2013} k!(k^3+3k^2+3k+1) - k!(k^2)$

$\displaystyle \sum_{k=1}^{2013} k!((k+1)^3) - k!(k^2) = \displaystyle \sum_{k=1}^{2013} (k+1)!((k+1)^2) - k!(k^2)$

$= (2!2^2 - 1!1^2) + (3!3^2 - 2!2^2) + (4!4^2 - 3!3^2) + ... + (2014!2014^2 - 2013!2013^2)$

$= 2014!2014^2 - 1!1^2$

$\implies n = 2014!2014^2$

Now exponent of $5$ in $n$ will be $402 + 80 + 16 + 3 = 501$

(formula used with proof: http://2000clicks.com/mathhelp/NumberTh10FactorialDivisors.aspx)

Now exponent of $2$ in $n$ will be greater than $1000.$

So exponent of $10$ in $n = 501 \implies$ no. of trailing zeros = $\boxed{501}$

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from math import factorial
n = 0
k_max = 2013

for k in range(1,k_max+1):
    n += factorial(k)*(k**3+2*k**2+3*k + 1)
n += 1

print 'Number of trailing zeros = %d' %(len(str(n)) - len(str(n).rstrip('0')))

Ans: 501