# Prove the equation using Newton's binom

• Lilia
In summary: Now add the first six entries (which include the two 5544s) and the last six entries (which also include two 5544s) and see that they are equal. You will have proved that ##\sum_{k=1}^6 k\, C(12,k) = \sum_{k=7}^{12} k \,C(12,k)##. But, of course, there is a bigger prize to be had: can you prove that ##\sum_{k=n}^{2n} k\, C(2n,k) = \sum_{k=1}^{n} k \,C(2n,n+k)##? If you
Lilia

## Homework Statement

prove that sum C(2n,k)*k = n*2^(2n-1); k=0,n

## Homework Equations

(x+y)^n = sum [C(n,k)*x^k*y^(n-k)]; k=0,n. // 1
in // 1, when y=1: (1+x)^n = sum [C(n,k)*x^k]; k=0,n // 2
in // 2, when x=1: 2^n = sum [C(n,k)]; k=0,n // 3

derivatives of both sides of (1+x)^n = sum [C(n,k)*x^k]; k=0,n:
n(1+x)^(n-1) = sum [k * C(n,k) * x^(k-1)]; k=0,n // 4
in // 4, when x=1: n*2^(n-1) = sum [k * C(n,k)]; k=0,n // 5

## The Attempt at a Solution

from // 5, i'll have
sum [C(2n,k)] = 2^(2n); k=0,2n

then i tried to divide this by 2:
sum [C(2n,k)] = 2^(2n)/2 = 2^(2n-1); k=0,n

but then i realized this can't be correct because there are
n+1 numbers in the range 0-n,
2n+1 numbers in the range 0-2n
so i can't really divide 2n+1 by two and get n+1

Lilia said:

## Homework Statement

prove that sum C(2n,k)*k = n*2^(2n-1); k=0,n

## Homework Equations

(x+y)^n = sum [C(n,k)*x^k*y^(n-k)]; k=0,n. // 1
in // 1, when y=1: (1+x)^n = sum [C(n,k)*x^k]; k=0,n // 2
in // 2, when x=1: 2^n = sum [C(n,k)]; k=0,n // 3

derivatives of both sides of (1+x)^n = sum [C(n,k)*x^k]; k=0,n:
n(1+x)^(n-1) = sum [k * C(n,k) * x^(k-1)]; k=0,n // 4
in // 4, when x=1: n*2^(n-1) = sum [k * C(n,k)]; k=0,n // 5

## The Attempt at a Solution

from // 5, i'll have
sum [C(2n,k)] = 2^(2n); k=0,2n

then i tried to divide this by 2:
sum [C(2n,k)] = 2^(2n)/2 = 2^(2n-1); k=0,n

but then i realized this can't be correct because there are
n+1 numbers in the range 0-n,
2n+1 numbers in the range 0-2n
so i can't really divide 2n+1 by two and get n+1

Hint: C(2n,n+k) = C(2n,n-k) and (n+k) = 2n - (n-k), so you can relate sum_{k=n..2n} k C(2n,k) = sum_{k=1..n} (n+k)C(2n,n+k) to sum_{k=1..n} k C(2n,k).

sum_{k=0..2n} k C(2n,k) = n*2^(2n)
if it was sum_{k=1..2n} k C(2n,k) = n*2^(2n), i'd just divide both sides by 2 and prove the equation

^it is
$$\sum_{k=0}^n k\,\mathrm{C}(2n,k)=\sum_{k=1}^n k\,\mathrm{C}(2n,k)$$

Lilia
yes...

but i don't understand why these are equal:
sum_{k=n..2n} k C(2n,k) = sum_{k=1..n} (n+k)C(2n,n+k)

Lilia said:
yes...

but i don't understand why these are equal:
sum_{k=n..2n} k C(2n,k) = sum_{k=1..n} (n+k)C(2n,n+k)

Who says they are equal?

Lilia said:

No, absolutely not. Saying that two things are related is not the same as saying they are equal.

is this correct?

sum {k=1,2n} c(2n,k) * x^k = (1+x)^(2n)
sum {k=1,2n} k * c(2n,k) * x^(k-1) = 2n * (1+x)^(2n-1)
sum {k=1,2n} c(2n,k) = 2^(2n-1), when x=1
n * 2^(2n-1) = [sum {k=1,2n} c(2n,k)] / 2 = sum {k=1,n} c(2n,k)

anyone?

^^only the second is correct though the sum should start at 0 to fit the others
in all k should start at 0
in the third and forth even though you had k * c(2n,k) in the second the k disappeared

sum {k=0,2n} c(2n,k) * x^k = (1+x)^(2n)
sum {k=0,2n} k * c(2n,k) * x^(k-1) = 2n * (1+x)^(2n-1)
sum {k=0,2n} k*c(2n,k) = 2^(2n-1), when x=1
n * 2^(2n-1) = [sum {k=0,2n} k*c(2n,k)] / 2 = sum {k=0,n} k*c(2n,k)

oh yes, that's true

but i don't get how these two are equal - [sum {k=0,2n} k*c(2n,k)] / 2 = sum {k=0,n} k*c(2n,k)

for the left part we have
c(2n,0)*0 + c(2n,1)*1 + ... + c(2n,n-1)*(n-1) + c(2n,n)*n + c(2n,n+1)*(n+1) + ... + c(2n,2n)*2n

having c(2n,k) = c(2n,2n-k), by summing all the coefficients of these c(2n,k) = c(2n,2n-k) i'll have that every term's coefficient is 2n, except for c(2n,n)*n.

after this, i don't think i can divide sum {k=0,2n} k*c(2n,k) by 2 and say it's equal to sum {k=0,n} k*c(2n,k)

Lilia said:
oh yes, that's true

but i don't get how these two are equal - [sum {k=0,2n} k*c(2n,k)] / 2 = sum {k=0,n} k*c(2n,k)

for the left part we have
c(2n,0)*0 + c(2n,1)*1 + ... + c(2n,n-1)*(n-1) + c(2n,n)*n + c(2n,n+1)*(n+1) + ... + c(2n,2n)*2n

having c(2n,k) = c(2n,2n-k), by summing all the coefficients of these c(2n,k) = c(2n,2n-k) i'll have that every term's coefficient is 2n, except for c(2n,n)*n.

after this, i don't think i can divide sum {k=0,2n} k*c(2n,k) by 2 and say it's equal to sum {k=0,n} k*c(2n,k)

For ##n=6## the sequence of values ##k \,C(12,k)## for ##k=1 \ldots 12## is
12, 132, 660, 1980, 3960, 5544, 5544, 3960, 1980, 660, 132, 12 .
We do, indeed, have ##\sum_{k=1}^6 k\, C(12,k) = \sum_{k=7}^{12} k \,C(12,k)= (1/2)\sum_{k=1}^{12} k\, C(12,k)##.

However, you need to prove that in general, for any ##n##.

Last edited:
but why are there two 5544's? we have only one c(12,6). for others there are equals, for example c(12,8) = c(12,4) etc.

Lilia said:
but why are there two 5544's? we have only one c(12,6). for others there are equals, for example c(12,8) = c(12,4) etc.

Do the calculation yourself: (1) compute C(12,k) for k = 1, 2, ..., 12; (2) multiply the entries by k = 1,2,..., 12, to get the sequence k*C(12,k) for k from 1 to 12.

It seems odd that the C's go up and then down again as k increases, while the entries k themselves continue to go up, but nevertheless their product goes up, then down and in a symmetric way. Until you have seen it happen you might not believe it; that is why a numerical example is helpful.

okay, i need to prove that sum {k=0,n} k*c(2n,k) = 1/2 * sum {k=0,2n} k*c(n/k)

for n=3,
for on left side i have ##\text{0*c(6,0) + 1*c(6,1) + 2*c(6,2) + 3*c(6,3) = 0 + 6 + 15 + 60 = 81}##
and on the right side ##\text{0*c(6,0) + 1*c(6,1) + 2*c(6,2) + 3*c(6,3) + 4*c(6,4) + 5*c(6,5) + 6*c(6,6)] = 81 + 60 + 30 + 6 = 177}##

since ##\text{c(n,k) = c(n,n-k)}##, here we can notice that the sum of the coefficients of c(2n,k) and c(2n,2n-k) is 2n but this isn't true for c(2n,n)
for example, ##\text{c(6,1) = c(6,5)}##, the sum of their coefficients: 1 + 5 = 6 = 2n, the same holds for the others but c(6,3) is unique

and also ##\frac {177} 2 ≠ 81##

Lilia said:
okay, i need to prove that sum {k=0,n} k*c(2n,k) = 1/2 * sum {k=0,2n} k*c(n/k)

for n=3,
for on left side i have ##\text{0*c(6,0) + 1*c(6,1) + 2*c(6,2) + 3*c(6,3) = 0 + 6 + 15 + 60 = 81}##
and on the right side ##\text{0*c(6,0) + 1*c(6,1) + 2*c(6,2) + 3*c(6,3) + 4*c(6,4) + 5*c(6,5) + 6*c(6,6)] = 81 + 60 + 30 + 6 = 177}##

since ##\text{c(n,k) = c(n,n-k)}##, here we can notice that the sum of the coefficients of c(2n,k) and c(2n,2n-k) is 2n but this isn't true for c(2n,n)
for example, ##\text{c(6,1) = c(6,5)}##, the sum of their coefficients: 1 + 5 = 6 = 2n, the same holds for the others but c(6,3) is unique

and also ##\frac {177} 2 ≠ 81##

Of course, ##\sum_{k=0}^n k C(2n,k) = \sum_{k=1}^n k C(2n,k)##, etc., and that is why I wrote the sums starting from 1 instead of 0. You get the required symmetry when you start at 1.

Beyond that, I don't know what else I can say without solving the problem for you. You need to actually evaluate numerically the terms ##k C(12,k)## to see what you get. But, I would suggest you do another case instead, so maybe look at ##n = 5## and thus look at the sequence ##k C(10,k)## for ##k = 1,2, \ldots, 10##. Do not leave the terms in symbolic form such as ##7 C(10,7)## for example; evaluate them numerically, so that you write ##840## instead of ##7 C(10,7).##

i calculated the second term wrong, so for n=3 the sequence of terms is
##\text{0 6 30 60 60 30 6}##
so it turns out that ##\text{n*c(2n,n) = (n+1)*c(2n,n+1)}## and i really wonder why and how

##\text{(n+1)*c(2n,n+1) = n*c(2n,n+1) + c(2n,n+1)}## but this doesn't get me anywhere

## 1. What is Newton's binomial equation?

Newton's binomial equation is a mathematical formula that can be used to expand a binomial raised to a power. It is expressed as (a + b)^n = Σ(nCk) * a^(n-k) * b^k, where a and b are constants, n is the power, and k is the term number.

## 2. How do you use Newton's binomial equation to prove an identity?

To prove an identity using Newton's binomial equation, you substitute the values of a and b into the equation and then simplify the resulting expression. If the simplified expression is equal to the original identity, then the identity is proven.

## 3. Can Newton's binomial equation be used for any power?

Yes, Newton's binomial equation can be used for any power, as long as the values of a and b are constants. However, the higher the power, the more terms will need to be expanded, making it more time-consuming and complex.

## 4. How is Newton's binomial equation related to the binomial theorem?

The binomial theorem is a general formula for expanding a binomial raised to any power. Newton's binomial equation is a specific case of the binomial theorem, where the power is a positive integer.

## 5. What are some real-life applications of Newton's binomial equation?

Newton's binomial equation has many practical applications in various fields such as physics, engineering, and finance. It can be used to model the growth of populations, calculate probabilities in statistics, and solve problems in kinematics and mechanics.

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