Self Doubt

Question

How to evaluate this quickly?

$\large\sum^{20}_{r=0}(-1)^r\binom{r+2}{r}\\OR\\\large\sum^{20}_{r=0}(-1)^r(r+2)(r+1)$

Answer 1

we can solve like this

Comments

@Prateek. Thank you.
Nice use of properties. :)

Answer 2

For odd values of r, the value will be negative, and for even, it will be positive.

So, break it in two series :

(0+2)(0+1) + (2+2)(2+1) + (4+2)(4+1) + ... +(20+2)(20+1) =

 

 $\sum_{t=0}^{10}(2t+2)(2t+1) = \sum_{t=0}^{10}(4t^{2}+6t + 2)$ = 1890

 

And :

 

-[(1+2)(1+1) +(3+2)(3+1) + (5+2)(5+1) +...+ (19+2)(19+1)] =

 

 $\sum_{t=1}^{10}(2t-1+2)(2t-1+1) = \sum_{t=1}^{10}(2t)(2t+1) = \sum_{t=1}^{10}(4t^{2}+2t)$  = 1644

 

Total sum = 1890 - 1644 = 246

Comments

@Gaurav . Thanks. Nice approach. :)
Just have some calculation mistake. The answer is 242.

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Yeah, wasn't sure about calculation part. My bad. :)

Answer 3

Suppose u want to get coefficient of $x^{20}$ in a series $\frac{1}{\left ( 1+x \right )\left ( 1-x \right )^{3}}$

Now

without direct formula $$\frac{1}{\left ( 1+x \right )}=1-x+x^{2}-x^{3}+...................(i)$$

and $$\frac{1}{\left ( 1-x \right )^{3}}=1+3x+\frac{3.4}{2!}x^{2}+\frac{3.4.5}{3!}x^{3}+............................(ii)$$

Now multiply series $(i)$ and $(ii)$

$1\times \frac{3.4.5.......22}{20!}-1\times \frac{3.4......21}{19!}+1\times \frac{3.4.....20}{18!}+........$

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direct formula $$\frac{1}{\left ( 1+x \right )}=1-x+x^{2}-x^{3}+...................(i)$$

and $$\frac{1}{\left ( 1-x \right )^{3}}=\sum_{0}^{\alpha }\binom{3+r-1}{r}...........................(ii)$$

but will be same calculation

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$\large\sum^{20}_{r=0}(-1)^r\binom{r+2}{r}$

$=\frac{1}{2}\sum^{20}_{r=0}(-1)^r\frac{(r+2)(r+1).r!}{r!2!}$

$=\frac{1}{2}\sum^{20}_{r=0}(-1)^r(r+2)(r+1)$

$\frac{1}{2}\left ( \left ( -1 \right )^{0}.2.1+(-1)^{1}3.2+(-1)^{2}4.3+...... \right )$

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Suppose question is

$\left ( \frac{1-x^{6}}{1-x} \right )^{4}$

Then formula will be

$\left ( 1-x^{6} \right )^{4}=\sum_{r=0}^{4}\binom{4}{r}\left ( -x^{6} \right )^{r}$

and $\left ( 1-x \right )^{-4}=\sum_{r=0}^{\alpha }\binom{4+r-1}{r}\left ( x \right )^{r}$

https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-5-exploring-the-infinite/part-b-taylor-series/session-99-taylors-series-continued/MIT18_01SCF10_Ses99b.pdf

Comments

Thanks, @Srestha :)
You explained the hidden part of my question. :P

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hidden part not your question

but hidden part of main ques calculation :)