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Question: The value of \[^n{C_1}{ + ^{n + 1}}{C_2}{ + ^{n + 2}}{C_3} + ...{ + ^{n + m - 1}}{C_m}\] is equal ...

The value of nC1+n+1C2+n+2C3+...+n+m−1Cm^n{C_1}{ + ^{n + 1}}{C_2}{ + ^{n + 2}}{C_3} + ...{ + ^{n + m - 1}}{C_m} is equal
(This question has multiple correct options)
A) m+nCn−1^{m + n}{C_{n - 1}}
B) m+nCn+1^{m + n}{C_{n + 1}}
C) mC1+m+1C2+m+2C3...+n+m−1Cn^m{C_1}{ + ^{m + 1}}{C_2}{ + ^{m + 2}}{C_3}...{ + ^{n + m - 1}}{C_n}
D) m+nCm−1^{m + n}{C_m} - 1

Explanation

Solution

Here we use the property of combinations that nCr+nCr+1=n+1Cr+1^n{C_r}{ + ^n}{C_{r + 1}}{ = ^{n + 1}}{C_{r + 1}}. Add and subtract the value nC0^n{C_0}from the given equation. Combine the pairs according to the property which will reduce a value in each step. Apply the property to the terms having the same subscript until we reach the end of the terms.

Complete step-by-step answer:
We are given the equation nC1+n+1C2+n+2C3+...+n+m−1Cm^n{C_1}{ + ^{n + 1}}{C_2}{ + ^{n + 2}}{C_3} + ...{ + ^{n + m - 1}}{C_m}.
Add and subtract the value nC0^n{C_0} from the equation
⇒nC0+nC1+n+1C2+n+2C3+...+n+m−1Cm−nC0………..…(1){ \Rightarrow ^n}{C_0}{ + ^n}{C_1}{ + ^{n + 1}}{C_2}{ + ^{n + 2}}{C_3} + ...{ + ^{n + m - 1}}{C_m}{ - ^n}{C_0} ………..… (1)
Now we know from the property that nCr+nCr+1=n+1Cr+1^n{C_r}{ + ^n}{C_{r + 1}}{ = ^{n + 1}}{C_{r + 1}}
Substituting the value of n=n,r=0n = n,r = 0 in the property
We can write nC0+nC0+1=n+1C0+1^n{C_0}{ + ^n}{C_{0 + 1}}{ = ^{n + 1}}{C_{0 + 1}}
Calculate the terms in subscript and superscript.
⇒nC0+nC1=n+1C1……..…(2){ \Rightarrow ^n}{C_0}{ + ^n}{C_1}{ = ^{n + 1}}{C_1} ……..… (2)
Substitute the value from equation (2) in equation (1)
⇒n+1C1+n+1C2+n+2C3+...+n+m−1Cm−nC0……….…(3){ \Rightarrow ^{n + 1}}{C_1}{ + ^{n + 1}}{C_2}{ + ^{n + 2}}{C_3} + ...{ + ^{n + m - 1}}{C_m}{ - ^n}{C_0} ……….… (3)
Now again we know from the property that nCr+nCr+1=n+1Cr+1^n{C_r}{ + ^n}{C_{r + 1}}{ = ^{n + 1}}{C_{r + 1}}
Substituting the value of n=n+1,r=1n = n + 1,r = 1
We can write n+1C1+n+1C1+1=n+1+1C1+1^{n + 1}{C_1}{ + ^{n + 1}}{C_{1 + 1}}{ = ^{n + 1 + 1}}{C_{1 + 1}}
Calculate the terms in subscript and superscript.
⇒n+1C1+n+1C2=n+2C2…...…(4){ \Rightarrow ^{n + 1}}{C_1}{ + ^{n + 1}}{C_2}{ = ^{n + 2}}{C_2} …...… (4)
Substitute the value from equation (4) in equation (3)
⇒n+2C2+n+2C3+...+n+m−1Cm−nC0{ \Rightarrow ^{n + 2}}{C_2}{ + ^{n + 2}}{C_3} + ...{ + ^{n + m - 1}}{C_m}{ - ^n}{C_0}
Similarly, n+2C2+n+2C3=n+3C3^{n + 2}{C_2}{ + ^{n + 2}}{C_3}{ = ^{n + 3}}{C_3}
So, the equation becomes
⇒n+3C3+...+n+m−1Cm−nC0{ \Rightarrow ^{n + 3}}{C_3} + ...{ + ^{n + m - 1}}{C_m}{ - ^n}{C_0}
Proceeding in this way we can write from the property.
n+m−1Cm−1+n+m−1Cm−1+1=n+mCm^{n + m - 1}{C_{m - 1}}{ + ^{n + m - 1}}{C_{m - 1 + 1}}{ = ^{n + m}}{C_m}
This will be the last term of the equation. So the equation becomes
⇒n+mCm−nC0{ \Rightarrow ^{n + m}}{C_m}{ - ^n}{C_0}
We know the value of nC0=1^n{C_0} = 1
Substitute the value of nC0=1^n{C_0} = 1 in the equation
⇒n+mCm−1{ \Rightarrow ^{n + m}}{C_m} - 1
Therefore, value of nC1+n+1C2+n+2C3+...+n+m−1Cm^n{C_1}{ + ^{n + 1}}{C_2}{ + ^{n + 2}}{C_3} + ...{ + ^{n + m - 1}}{C_m} is n+mCm−1^{n + m}{C_m} - 1

Note: Students might try to open up each combination using the formula nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}which will result in a sum which is difficult to calculate. It is advised to use the property which converts the sum of two terms into one term and we move in the same manner until we reach the end.Students are advised to perform the first two sums of combinations step by step in order to understand the pattern.