Just elegancy required! 4

Algebra Level 4

$\displaystyle \sum_{n=1}^{9} \frac{n \cdot 2^{n}}{(n+2)!} = \frac{A}{A+4}$ Find $A$ .

The answer is 155921.

2 solutions

Rishabh Jain
Feb 16, 2016

$\Large\displaystyle \sum_{n=1}^{9} \dfrac{((n+\color{#D61F06}{2})-\color{#D61F06}{2} )\cdot 2^{n}}{(n+2)!}$ $\Large=\displaystyle \sum_{n=1}^{9} \dfrac{2^{n}}{(n+1)!} -\dfrac{2^{n+1}}{(n+2)!}\\$ $\large\color{#302B94}{\mathfrak{(A~Telescopic~Series)} }\\$ $\Large =1-\dfrac{2^{10}}{11!}\\$ $\Large=1-\dfrac{4}{(\color{forestgreen}{\frac{11!}{2^8}-4})+4}\\$ Hence, $\huge A=\boxed{\color{forestgreen}{155921}}$

$\text{Note }\frac{A}{A+4}\text{(i.e RHS) can be written as}\\ 1-\frac{4}{\color{forestgreen}{A}+4} \text{ and hence on comparison we get}\\ A=\frac{11!}{2^8}-4$

Great observation at last .....really cool

Mohit Gupta - 5 years, 3 months ago

Nice question !!!!

Akshat Sharda - 5 years, 3 months ago

U r good in it bravo!!!!!!!

aishwarya dadhich - 5 years, 3 months ago

Nice observation.

Niranjan Khanderia - 5 years, 3 months ago

Thanks.. :-}

Rishabh Jain - 5 years, 3 months ago

Same way. Another nice solution of yours bro!

Shreyash Rai - 5 years, 3 months ago

$\huge\color{#302B94}{\mathbb{THANKS}}$

Rishabh Jain - 5 years, 3 months ago

How to get this telescopic series? I mean how to solve telescopic series with factorials.

Anshuman Singh Bais - 5 years, 3 months ago

For this particular question?? $\displaystyle \sum_{n=1}^{9} \dfrac{2^{n}}{(n+1)!} -\dfrac{2^{n+1}}{(n+2)!}\\$ $\small{(\dfrac{2}{2!}-\not{\dfrac{2^2}{3!}})+(\not{\dfrac{2^2}{3!}}-\not{\dfrac{2^3}{3!}})+(\not{\dfrac{2^3}{3!}}\cdots\cdots -\not{\dfrac{2^9}{10!}})+(\not{\dfrac{2^{9}}{10!}}-\dfrac{2^{10}}{11!}}$ $=1-\dfrac{2^{10}}{11!}$

Rishabh Jain - 5 years, 3 months ago

$\sum_{n=0}^{9}\frac{2^n}{(n+1)!}-\sum_{n=0}^{9}\frac{2^{n+1}}{(n+2)!} \\ =\sum_{n=0}^{9}\frac{2^n}{(n+1)!}-\sum_{r=1}^{10}\frac{2^{r}}{(r+1)!} = \frac2{2!} - \frac{2^{10}}{11!}$

Here, $r=n+1$ in the second summation.

shaurya gupta - 5 years, 3 months ago

Spider Webs
Mar 26, 2016

I sort of cheated on this one. Since we were told the numerator would be 4 less than the denominator, I found the lowest common denominator for (n+2)! from n = 1 to n = 9, or 3! through 11!, which would of course be 11!, since it includes the previous factorials.

The prime factorization for 11! is ${2^8} * {3^4} * {5^2} * 7 * 11$ . Applying Legendre's Theorem , every term in the series would have enough 2s in the numerator to cancel the 2s in the denominator. (The maximum number of 2s in the denominator relative to (n+2) would occur when (n+2) is a power of 2, resulting in (n+1) 2s. However, there are n 2s in ${2^n}$ , and n would necessarily be a multiple of 2 if (n+2) is a power of 2, which would provide the remaining 2 required to cancel the (n+1) 2s in the denominator.)

Since every 2 would cancel from each fraction, the LCD would be 11! without any 2s. So... LCD: ${3^4} * {5^2} * 7 * 11$ = 155925 = A + 4. Hence A = 155921. There is no possibility that $\frac{A}{A + 4}$ represents a non-simplified version of the fraction, since the difference between A + 4 and A is 4, which is made up of only 2s, which is not a factor of A.

Just elegancy required! 4

The answer is 155921.

2 solutions

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