Find value of C0+(C0+C1)+(C0+C1+C2)+(C0+C1+C2+C3).

[vkash]

find value of C0+(C0+C1)+(C0+C1+C2)+(C0+C1+C2+C3)...

Cr represent ⁿC_r
find the value of C0+(C0+C1)+(C0+C1+C2)+(C0+C1+C2+C3)...(C0+C1+C2+C3+...Cn)

How i did it
C0+C1+C2+...Cn=(1+1)ⁿ
so
C0=(1+1)⁰
C0+C1=(1+1)¹ ( here n is 1)
C0+C1+C2=(1+1)² (here n is 2)
.
.
.

so the required question is changed into following
2⁰+2¹+2²+2³+...+2ⁿ
that's Geometric progression
so it should equal to 2ⁿ-1

where i have done it wrong

 

[eumyang]

Cr represent _nC_r
find the value of C0+(C0+C1)+(C0+C1+C2)+(C0+C1+C2+C3)...(C0+C1+C2+C3+...Cn)

How i did it
C0+C1+C2+...Cn=(1+1)ⁿ
so
C0=(1+1)⁰
C0+C1=(1+1)¹ ( here n is 1)
C0+C1+C2=(1+1)² (here n is 2)

Your notation is ambiguous. If Cr = _nC_r, then I would think that
C0 = _nC₀ = 1,
C0 + C1 = _nC₀ + _nC₁ = 1 + n,
C0 + C1 + C2 = _nC₀ + _nC₁ + _nC₂ = 1 + n + n(n+1)/2.

It looks like when you were finding
C0+(C0+C1)+(C0+C1+C2)+(C0+C1+C2+C3)...(C0+C1+C2+C3+...Cn)
you were actually finding
₀C₀ + (₁C₀ + ₁C₀) + (₂C₀ + ₂C₀ + ₂C₀) + ... + (_nC₀ + _nC₀ + _nC₀ + ... + _nC₀).

So which one are you looking for, exactly?

 

[vkash]

Your notation is ambiguous. If Cr = _nC_r, then I would think that
C0 = _nC₀ = 1,
C0 + C1 = _nC₀ + _nC₁ = 1 + n,
C0 + C1 + C2 = _nC₀ + _nC₁ + _nC₂ = 1 + n + n(n+1)/2.

It looks like when you were finding
C0+(C0+C1)+(C0+C1+C2)+(C0+C1+C2+C3)...(C0+C1+C2+C3+...Cn)
you were actually finding
₀C₀ + (₁C₀ + ₁C₀) + (₂C₀ + ₂C₀ + ₂C₀) + ... + (_nC₀ + _nC₀ + _nC₀ + ... + _nC₀).

So which one are you looking for, exactly?

my question was correct.
You did not answer the question but you answer you have solved my problem. that is always ⁿC_r. I take different values of n. that's what i was doing wrong.
Thanks Bcz you put out difference in my answer and question.
Thanks friend.

 

[Ray Vickson]

Assuming you mean C(k) = _nC_k for k = 0, 1, 2, ..., n, your sum, S, can be expressed as
$$\begin{array}{l}S = (n+1)C(0) + n C(1) + (n-1) C(2) + \cdots + C(n) \\ \mbox{ } = (n+1)[C(0) + C(1) + \cdots + C(n)] - [C(1) + 2C(2) + \cdots + nC(n)], \end{array}$$
and this last sum can be computed (do you see how?)

RGV