Another big blow summation

Algebra Level 4

$\large \sum _{r=1}^{20} (-1)^r \frac {r^2 + r + 1}{r!} = \frac a{b!} - 1$

The equation above holds true for some coprime positive integers $a$ and $b$ .

Find $a + b$ .

The answer is 41.

2 solutions

Chew-Seong Cheong
Nov 28, 2016

Expanding on @Rishabh Cool's solution:

$\begin{aligned} S & = \sum_{r=1}^{20} (-1)^r \frac {r^2+r+1}{r!} \\ & = \sum_{r=1}^{20} (-1)^r \left(\frac {r^2}{r!} + \frac {r+1}{r!} \right) \\ & = \sum_{r=1}^{20} (-1)^r \left(\frac {r^2}{r(r-1)!} + \frac {r+1}{r!} \right) \\ & = \sum_{r=1}^{20} (-1)^r \left(\frac r{(r-1)!} + \frac {r+1}{r!} \right) \\ & = - \frac 1{0!} - \frac 2{1!} + \frac 2{1!} + \frac 3{2!} - \frac 3{2!} - \frac 4{3!} + ... + \frac {20}{19!} + \frac {21}{20!} \\ & = \frac {21}{20!} - 1 \end{aligned}$

$\implies a + b = 21+20 = \boxed{41}$

Both @Chew-Seong Cheong , and @Rishabh Cool

how you think of breaking the term in two terms?

Do you apply any logic in breaking the terms or just by hit and trial method?

Priyanshu Mishra - 4 years, 6 months ago

Don't get what you mean. I have added new second and third steps to explain it.

Chew-Seong Cheong - 4 years, 6 months ago

I fully understand your steps, no doubt in it.

I am asking in general that how you write the very first step i.e to break the given terms in difference of two terms.

$\large \sum_{r=1}^{20}{(-1)^r}\left(\frac r{(r-1)!} + \frac {r+1}{r!} \right)$

How you thought of breaking into these terms?

Priyanshu Mishra - 4 years, 6 months ago

@Priyanshu Mishra – To make two terms to have the same degree in the nominator and denominator. How else but to divide $r^2$ by $r!$ . A lot is try and error. You have to do more problems to get it.

Chew-Seong Cheong - 4 years, 6 months ago

@Chew-Seong Cheong – Thanks i got it.

Priyanshu Mishra - 4 years, 6 months ago

@Priyanshu Mishra – Chew, see my comment to Rishabh. Thanks, Ed Gray

Edwin Gray - 2 years, 8 months ago

@Priyanshu Mishra – You should also check out this problem .

Chew-Seong Cheong - 4 years, 6 months ago

Rishabh Jain
Nov 28, 2016

The sum is : $\large\sum_{r=1}^{20}(-1)^r\dfrac{\overbrace{\underbrace{r^2+r+1}}^{(r+1)+(r^2) }}{r!}$

$=\sum_{r=1}^{20}(-1)^r\left\{\dfrac{r+1}{r!}+\dfrac{r}{(r-1)!}\right\}$

Writing few terms we see this is a $\small{\color{#0C6AC7}{\text{telescopic series}}}$ .

$=\dfrac{21}{20!}-1$

$\therefore 21+20=\boxed{41}$

I see no advantage in breaking up the term r^2 + r + 1. Either way, when r = 1, the first term is [(-1)^1]*(1^2 + 1 + 1)/1! = -3. Using the above break up, we have [(-1)^1]{2/1! + 1/0!} = -3. The second term is +2.5. Adding the first few terms, we have: -3 + 3.5 - 2.1666666 +.875 -.25833333333 + .059722222 -.011309523 + ….. The sum approaches -1. If this is so, a must equal 0, and b could be anything Ed Gray.

Edwin Gray - 2 years, 8 months ago

But the sum is not -1, it is just close to it. I have checked it with Wolfram Alpha (see this link ). The sum is $\dfrac {21}{20!} - 1 \approx -1$ .

Chew-Seong Cheong - 2 years, 8 months ago

Chew, you are absolutely correct. Nice solution; regards, Ed

Edwin Gray - 2 years, 8 months ago

Another big blow summation

The answer is 41.

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