Calcgebra

Calculus Level 5

First, compute

$\mathfrak{S}= \displaystyle \sum_{e=1}^{5} \sum_{d=1}^{e} \sum_{c=1}^{d} \sum_{b=1}^{c} \sum_{a=1}^{b} 1$ .

If the above sum $\mathfrak{S} =\zeta ^{3} + 1$ , with $\zeta \in \mathbb {N}$ then evaluate

$\mathfrak{I} = \displaystyle \int_{0}^{\pi/4} \tan^{\zeta^{3}} x \sec^{4} x \, dx$

Finally, if $\mathfrak{I}$ can be represented in the simple form $\dfrac{a}{b}$ , where $a, b$ are coprime positive integers, find $a+b$ .

Nota Bene: NO CALCULATORS!

Bonus: Can you find a way to compute $\mathfrak{S}$ efficiently?

Inspiration

The answer is 8191.

1 solution

A Former Brilliant Member
Apr 12, 2016

Solution only for calculating the sum,
$S = \displaystyle \sum_{e=1}^{5} \sum_{d=1}^{e} \sum_{c=1}^{d} \sum_{b=1}^{c} \sum_{a=1}^{b}1$
$S =\displaystyle \sum_{e=1}^{5} \sum_{d=1}^{e} \sum_{c=1}^{d} \sum_{b=1}^{c} \dbinom{b}{1}$
$S =\displaystyle \sum_{e=1}^{5} \sum_{d=1}^{e} \sum_{c=1}^{d} \dbinom{c+1}{2}$
$S =\displaystyle \sum_{e=1}^{5} \sum_{d=1}^{e} \dbinom{d+2}{3}$
$S = \displaystyle \sum_{e= 1}^{5} \dbinom{e+3}{4}$
$S = \displaystyle \dbinom{5+4}{5} = \dbinom{9}{5} = 126$
This can very easily be generalised to n variables
I used,
$\displaystyle \sum_{m=0}^{n} \dbinom{m} {k} = \dbinom{n+1}{k+1}$
This is known as the hockey stick identity and can be easily derived using $\dbinom{n}{r} + \dbinom{n}{r-1} = \dbinom{n+1}{r}$

Similar summation :
Summation over 7 variables

Thanks for your solution! I learned something! :)

Hobart Pao - 5 years, 2 months ago

Calcgebra

The answer is 8191.

1 solution

0 pending reports