Question

fun question about binom identities

Answer 1

@imqwerty

Answer 2

@Astrophysics does your text-box kinda freeze when you post a reply?

Answer 3

Sometimes

Answer 4

Wait, hold on, the question is wrong I think.

Answer 5

This is again wrong. lol\[\sum_{k=1}^{2017}(-1)^{k+1}(7k+1) \binom{2016}{k-1}\]

Answer 6

i am treating 2017 as n
okay we can break it like this-

\[\large 1\left( \sum_{k=1}^{n} (-1)^{k} ~^{n}C_{k-1}\right)+7\left( \sum_{k=1}^{n}(-1)^k (n)~^{n-1}C_{k-2} \right)\]

now we jst have to apply binomial identities which we get from calculus
i am writing this 3rd time ;-;

Answer 7

we know that
\(^nC_0-~^nC_1+~^nC_2-........(-1)^n ~^nC_n=0\) <--identity

and this part of this eq-\[\large 1\left( \sum_{k=1}^{n} (-1)^{k} ~^{n}C_{k-1}\right)\]
is this->\(-^nC_0+~^nC_1-~^nC_2+....(-1)^{2017} ~^{2017}C_{2016}\)

making some adjustments and using that identity we get its value as-> -1

Answer 8

awww D: i made a typo up there
heres the correction- http://prntscr.com/9m2d6x

Answer 9

okay for 2nd part-\[7\left( \sum_{k=2}^{n}(-1)^k (n)~^{n-1}C_{k-2} \right)\]\[7n\left( \sum_{k=2}^{n}(-1)^k ~^{n-1}C_{k-2} \right)\]
that series inside the bracket will go liek this->
\(^{n-1}C_0-~^{n-1}C_1+~^{n-1}C_2-..........(-1)^n ~^{n-1}C_{n-2}\)
we just have to use that same identity here
this summation of series will be easily obtained as-> -1

Answer 10

summation of inside series is -1
but we gotta multiply it with 7n which is outside where n=2017

so we get-> \(7(2017)(-1)=-14119\) this number makes me feel something is wrong but anyways ima continue

so we got values of both parts
adding them we get-> \(-1-14119=-14120\)

Answer 11

It is zero

Answer 12

\[\begin{align*}S&=\sum_{k=1}^{2017}(-1)^{k+1}(7k+1)\binom{2016}{k-1}\\[1ex]
&=\sum_{k=0}^{2016}(-1)^{k+2}(7k+8)\binom{2016}k\\[1ex]
&=\sum_{k=0}^{2016}(-1)^k(7k+8)\binom{2016}k\\[1ex]
&=7\sum_{k=0}^{2016}(-1)^kk\binom{2016}k+8\sum_{k=0}^{2016}(-1)^k\binom{2016}k\\[1ex]

\end{align*}\]
Recall the binomial theorem:
\[\sum_{k=0}^{2016}\binom{2016}k(-1)^k=\sum_{k=0}^{2016}\binom{2016}k(-1)^k1^{2016-k}=(1-1)^{2016}=0\]leaving us with
\[S=7\sum_{k=0}^{2016}(-1)^k k\binom{2016}k\]
From the binomial theorem, you have that
\[(x-1)^{2016}=\sum_{k=0}^{2016}\binom{2016}kx^k(-1)^{2016-k}=\sum_{k=0}^{2016} \binom{2016}k x^k(-1)^k\]Differentiating once with respect to \(x\), you have
\[2016(x-1)^{2015}=\sum_{k=0}^{2016}\binom{2016}kkx^{k-1}(-1)^k\]and notice that \(S\) is obtained with \(x=1\) again, so
\[S=2016(1-1)^{2015}=0\]