Another Binomial Madness

Calculus Level 5

$(1+x)^n=C_0+C_1x + C_2x^2 + \cdots + C_rx^r + \cdots + C_n x^n$

Let $n$ be a positive integer such that the above shows an algebraic identity such that the variables $C_0, C_1, \ldots, C_n$ are independent of $x$ .

$\large \dfrac{C_0}{1} - \dfrac{C_1}{8} + \dfrac{C_2}{15} - \cdots +(-1)^n \dfrac{C_n}{7n+1}$

If the closed form of the expression above can be expressed as $\dfrac{a \cdot 7^{n+b} \cdot (n+c)!}{1\cdot8\cdot15\cdots(7n+1)} \; ,$ where $a,b$ and $c$ are non-negative integers, find $a+b+c$ .

Clarification :
$!$ denotes the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$ .

Click here for Part I of the problem.

The answer is 1.

1 solution

A Former Brilliant Member
Apr 10, 2016

$(1-x^{7})^{n}= \displaystyle \sum_{r=0}^{n} \dbinom{n}{r}(-1)^{r}x^{7r}$
$I = \displaystyle \int_{0}^{1} (1-x^{7})^{n}dx = \displaystyle \sum_{r=0}^{n} \int_{0}^{1} \dbinom{n}{r}(-1)^{r}x^{7r}dx$
$\therefore I = \displaystyle \sum_{r=0}^{n} \dfrac{\dbinom{n}{r}(-1)^{r}}{7r+1}$
$I = \displaystyle \int_{0}^{1}(1-x^{7})^{n} dx$
Substitute $x = t^{\frac{1}{7}}$ . The integral transforms into,
$I =\dfrac{\beta(\frac{1}{7},n+1)}{7}$ where $\beta$ deotes the beta function .
$I = \dfrac{\Gamma(\frac{1}{7})\Gamma(n+1)}{7\Gamma(\frac{7n+8}{7})} = \dfrac{7^{n}(n)!}{1\cdot 8 \cdot 15 \ldots \cdot (7n+1)}$
Comparing we get,
$a = 1, b = c = 0$
$a + b + c = 1$

I didn't understood the step just before where you wrote : Comparing we get .. I mean how You evaluated that gamma function?(I am still learning them).

Anurag Pandey - 4 years, 8 months ago

Did by induction

A Former Brilliant Member - 3 years, 4 months ago

Another Binomial Madness

Click here for Part I of the problem.

The answer is 1.

1 solution

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