Question

The value of \[^n{C_1}{ + ^{n + 1}}{C_2}{ + ^{n + 2}}{C_3} + ...{ + ^{n + m - 1}}{C_m}\] is equal
(This question has multiple correct options)
A) \[^{m + n}{C_{n - 1}}\]
B) \[^{m + n}{C_{n + 1}}\]
C) \[^m{C_1}{ + ^{m + 1}}{C_2}{ + ^{m + 2}}{C_3}...{ + ^{n + m - 1}}{C_n}\]
D) \[^{m + n}{C_m} - 1\]

Answer 1

Hint:Here we use the property of combinations that \[^n{C_r}{ + ^n}{C_{r + 1}}{ = ^{n + 1}}{C_{r + 1}}\]. Add and subtract the value \[^n{C_0}\]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 \[^n{C_1}{ + ^{n + 1}}{C_2}{ + ^{n + 2}}{C_3} + ...{ + ^{n + m - 1}}{C_m}\].
Add and subtract the value \[^n{C_0}\] from the equation
\[{ \Rightarrow ^n}{C_0}{ + ^n}{C_1}{ + ^{n + 1}}{C_2}{ + ^{n + 2}}{C_3} + ...{ + ^{n + m - 1}}{C_m}{ - ^n}{C_0} ………..… (1)\]
Now we know from the property that \[^n{C_r}{ + ^n}{C_{r + 1}}{ = ^{n + 1}}{C_{r + 1}}\]
Substituting the value of \[n = n,r = 0\] in the property
We can write \[^n{C_0}{ + ^n}{C_{0 + 1}}{ = ^{n + 1}}{C_{0 + 1}}\]
Calculate the terms in subscript and superscript.
\[{ \Rightarrow ^n}{C_0}{ + ^n}{C_1}{ = ^{n + 1}}{C_1} ……..… (2)\]
Substitute the value from equation (2) in equation (1)
\[{ \Rightarrow ^{n + 1}}{C_1}{ + ^{n + 1}}{C_2}{ + ^{n + 2}}{C_3} + ...{ + ^{n + m - 1}}{C_m}{ - ^n}{C_0} ……….… (3)\]
Now again we know from the property that \[^n{C_r}{ + ^n}{C_{r + 1}}{ = ^{n + 1}}{C_{r + 1}}\]
Substituting the value of \[n = n + 1,r = 1\]
We can write \[^{n + 1}{C_1}{ + ^{n + 1}}{C_{1 + 1}}{ = ^{n + 1 + 1}}{C_{1 + 1}}\]
Calculate the terms in subscript and superscript.
\[{ \Rightarrow ^{n + 1}}{C_1}{ + ^{n + 1}}{C_2}{ = ^{n + 2}}{C_2} …...… (4)\]
Substitute the value from equation (4) in equation (3)
\[{ \Rightarrow ^{n + 2}}{C_2}{ + ^{n + 2}}{C_3} + ...{ + ^{n + m - 1}}{C_m}{ - ^n}{C_0}\]
Similarly, \[^{n + 2}{C_2}{ + ^{n + 2}}{C_3}{ = ^{n + 3}}{C_3}\]
So, the equation becomes
\[{ \Rightarrow ^{n + 3}}{C_3} + ...{ + ^{n + m - 1}}{C_m}{ - ^n}{C_0}\]
Proceeding in this way we can write from the property.
\[^{n + m - 1}{C_{m - 1}}{ + ^{n + m - 1}}{C_{m - 1 + 1}}{ = ^{n + m}}{C_m}\]
This will be the last term of the equation. So the equation becomes
\[{ \Rightarrow ^{n + m}}{C_m}{ - ^n}{C_0}\]
We know the value of \[^n{C_0} = 1\]
Substitute the value of \[^n{C_0} = 1\] in the equation
\[{ \Rightarrow ^{n + m}}{C_m} - 1\]
Therefore, value of \[^n{C_1}{ + ^{n + 1}}{C_2}{ + ^{n + 2}}{C_3} + ...{ + ^{n + m - 1}}{C_m}\] is \[^{n + m}{C_m} - 1\]

Note:Students might try to open up each combination using the formula \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]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.