Question

Plzzz help!
Find the value of C(7,0) + C(7,1) + C(7,1) + C(7,2) +...+ C(7, 6) + (7, 6)

Answer 1

C(7,0) + C(7,1) + C(7,1) + C(7,2) +...+ C(7, 6) + (7, 6) ??
define C is it constant ??

Answer 2

What exactly is the pattern?

Answer 3

Is it \[
\sum_{k=0}^n2\binom {n}{k}
\]?

Answer 4

Because that is just \[

\sum_{k=0}^n2\binom {n}{k}=
2\sum_{k=0}^n\binom {n}{k}=2\times 2^n=2\times 2^7=256
\]

Answer 5

But somehow you're missing terms.

Answer 6

I mean to be really picky what you have is: \[
\left[\sum_{k=0}^n2\binom {n}{k}\right]-\binom 70 -2\binom 77=\left[\sum_{k=0}^n2\binom {n}{k}\right]-3=256-3=253
\]

Answer 7

mmmm the easy way !
C(7,0) + C(7,1) + C(7,1) + C(7,2) + C(7,3)+ C(7,4)+ C(7,5)+ C(7, 6) + (7, 6)
\[=(C(7*n+2),\frac{ n(n+1) }{ 2 } +6)\]
such that n=7.

Answer 8

thank u all
but i think i am getting even more confused
Do u know this formula:
\[^{n} C_{r} =\frac{ n! }{ r!(n-r)! }\]
I think we are supposed to use this formula to solve this question.
I have been trying soo hard but still I am not getting to anything almost 5 or 6 pages full of rough work
I will try solving using the hint by @ikram002p
so thanx @ikram002p

Answer 9

i m sorry but i completely do not understand using your method @wio
i knw u r using summation to solve but i do not knw tht method so thts why but ill give it a try as well so thnx to you @wio as well :)

Answer 10

My method comes from the binomial theorem: \[
\sum_{k=0}^{n}\binom nk a^kb^{n-k}=(a+b)^n
\]When you let \(a=1,b=1\) then: \[
\sum_{k=0}^{n}\binom nk 1^k1^{n-k}=(1+1)^n
\]Which simplifies to \[

\sum_{k=0}^{n}\binom nk=2^n
\]

Answer 11

Remember that: \[
\sum_{k=0}^{n}\binom nk =\binom n0+\binom n1 +\binom n2+\ldots +\binom n{n-1}+\binom nn
\]