Question

Prove that \[^n{C_{r - 1}}{ + ^n}{C_r}{ = ^{n + 1}}{C_r}\].

Answer 1

Hint – In order to solve this problem you just need to know the formula \[^a{C_b} = \dfrac{{a!}}{{b!(a - b)!}}\] then this problem will be solved.

Complete step-by-step answer:
We have to prove, \[^n{C_{r - 1}}{ + ^n}{C_r}{ = ^{n + 1}}{C_r}\]. ……………………………………(1)
We know the formula that \[^a{C_b} = \dfrac{{a!}}{{b!(a - b)!}}\]
As we know that a! = (a-3)!(a-2)(a-1)a
With the help of this concept we just have to solve LHS and get the value of RHS.
So, we do
\[
  ^n{C_{r - 1}}{ + ^n}{C_r} = \dfrac{{n!}}{{(r - 1)!(n - r + 1)!}} + \dfrac{{n!}}{{r!(n - r)!}} \\
  ^n{C_{r - 1}}{ + ^n}{C_r} = \dfrac{{n!}}{{(r - 1)!(n - r + 1)(n - r)!}} + \dfrac{{n!}}{{r(r - 1)!(n - r)!}} \\
  ^n{C_{r - 1}}{ + ^n}{C_r} = \dfrac{{n!}}{{(r - 1)!(n - r)!}}\left[ {\dfrac{1}{{n - r + 1}} + \dfrac{1}{r}} \right] \\
  ^n{C_{r - 1}}{ + ^n}{C_r} = \dfrac{{n!}}{{(r - 1)!(n - r)!}}\left[ {\dfrac{{n + r - r + 1}}{{nr - {r^2} + r}}} \right] \\
  ^n{C_{r - 1}}{ + ^n}{C_r} = \dfrac{{n!}}{{(r - 1)!(n - r)!}}\left[ {\dfrac{{n + 1}}{{(n - r + 1)r}}} \right] \\
  ^n{C_{r - 1}}{ + ^n}{C_r} = \dfrac{{n!(n + 1)}}{{(r - 1)!(n - r)!(n - r + 1)r}} = \dfrac{{(n + 1)!}}{{r!(n + 1 - r)!}} \\
\]
If we see carefully then we can recognize that \[\dfrac{{(n + 1)!}}{{r!(n + 1 - r)!}}\] is nothing but \[^{n + 1}{C_r}\].
If we expand \[^{n + 1}{C_r}\] then we will be getting \[\dfrac{{(n + 1)!}}{{r!(n + 1 - r)!}}\]. ……………………….(2)
So, from (1)and (2) it is proved that \[^n{C_{r - 1}}{ + ^n}{C_r}{ = ^{n + 1}}{C_r}\].

Note – When you are asked to prove this then you have to simply expand the LHS and solve as I have done above and get the value of RHS. Better you calculate the value of RHS so that you are sure that your solution is correct. Proceeding in this way will solve your problem.