# find the sum of series nC1+(n+1)C2+(n+2)C3+.........+(n+r-1)Crplease could u help me how to proceed in thesetype of problems

Arun Kumar IIT Delhi
10 years ago
$\sum_{k=1}^{r}_{k}^{n+k-1}\textrm{C}\\=>\sum_{k=1}^{r}_{n-1}^{n+k-1}\textrm{C}\\ =>_{n}^{n}\textrm{C}+\sum_{k=1}^{r}_{n-1}^{n+k-1}\textrm{C}-_{n}^{n}\textrm{C}\\ since\,_{n}^{n}\textrm{C}+_{n}^{n-1}\textrm{C}=_{n+1}^{n}\textrm{C}\\ =>_{n}^{n+r}\textrm{C}-1$

Arun Kumar
IIT Delhi
Arun Kumar IIT Delhi
10 years ago

$since\,_{n}^{n}\textrm{C}+_{n-1}^{n}\textrm{C}=_{n}^{n+1}\textrm{C}\\$
Saumya
13 Points
6 years ago
N(C1+C2+......Cr)+( (2-1)C2+(3-1)C3......) SUMMATION Cr + SUMMATION rCrSUMMATION Cr is 2^nAnd to solve rCr write expansion of (1+x)^n and differentiate and then put x=1
4 years ago
Dear student,

nC1 + n+1C2 + n+2C3 + .........+ n+r–1Cr
= nC0 + nC1 + n+1C2 + n+2C3 + .........+ n+r-1Cr – nC      {adding and subtracting nC0}
= n+1C1 + n+1C2 + n+2C3 + .........+ n+r-1Cr – 1
{Since we know, nCr + nCr+1 = n+1Cr+1}
= n+2C2 + n+2C3 + .........+ n+r-1Cr – 1
With each term we can apply the same formula
Hence, the condensed result will be
= n+rCr – 1 = n+rCn – 1

Thanks and regards,
Kushagra