Question

For any positive integer $n,(a+b)^n$ is equal to

• A. ${ }^n C_1 a^n+{ }^n C_2 a^{n-1} b+{ }^n C_3 a^{n-2} b^2+\ldots \ldots \ldots+{ }^n C_{n+1} b^n$
• B. ${ }^n C_0 b^n+{ }^n C_1 b^{n-3} a+{ }^n C_2 b^{n-5} a^2+\ldots . .+{ }^n C_n a^n$
• C. ${ }^n C_0 a^n+{ }^n C_1 a^{n-1} b+{ }^n C_2 a^{n-2} b^2+\ldots \ldots +{ }^n C_{n-1} a b^{n-1}+{ }^n C_n b^n $
• D. None of the above

Answer 1

Answer: (C)

The proof of this result is given below
Let the given statement be
$ P(n):(a+b)^n={ }^n C_0 a^n+{ }^n C_1 a^{n-1} b+{ }^n C_2 a^{n-2} b^2+ $
$ \ldots+{ }^n C_{n-1} a \cdot b^{n-1}+{ }^n C_n b^n$
For $n=1$, we have
$P(1):(a+b)^1={ }^1 C_0 a^1+{ }^1 C_1 b^1=a+b$
Thus, $P(1)$ is true.
Suppose $P(k)$ is true for some positive integer $k$, i.e.,
$(a+b)^k={ }^k C_0 a^k+{ }^k C_1 a^{k-1} b+{ }^k C_2 a^{k-2} b^2+\ldots+{ }^k C_k b^k \ldots \text { (i) }$
We shall prove that $P(k+1)$ is also true, i.e.,
$(a+b)^{k+1}={ }^{k+1} C_0 a^{k+1}+{ }^{k+1} C_1 a^k b+{ }^{k+1} C_2 a^{k-1} b^2+ $
$\cdots+{ }^{k+1} C_{k+1} b^{k+1}$
Now, $(a+b)^{k+1}=(a+b)(a+b)^k$
$=(a+b)\left({ }^k C_0 a{ }^k+{ }^k C_1 a^{k-1} b+{ }^k C_2 a^{k-2} b^2+\right. $
$\left.\ldots+{ }^k C_{k-1} a b^{k-1}+{ }^k C_k b^k\right)[\text { from Eq. (i) }]$
$={ }^k C_0 a{ }^{k+1}+{ }^k C_1 a^k b+{ }^k C_2 a{ }^{k-1} b^2+\ldots+{ }^k C_{k-1} a^2 b^{k-1}$
$ +{ }^k C_k a b^k+{ }^k C_0 a^k b+{ }^k C_1 a^{k-1} b^2+{ }^k C_2 a^{k-2} b^3+ $
$ \ldots+{ }^k C_{k-1} a b^k+{ }^k C_k b^{k+1} \text { (by actual multiplication) }$
$={ }^k C_0 a^{k+1}+\left({ }^k C_1+{ }^k C_0\right) a^k b+\left({ }^k C_2+{ }^k C_1\right) a^{k-1} b^2+ $
$ \ldots+\left({ }^k C_k+{ }^k C_{k-1}\right) a b^k+{ }^k C_k b^{k+1} \text { (grouping like terms) }$
$ ={ }^{k+1} C_0 a^{k+1}+{ }^{k+1} C_1 a^k b+{ }^{k+1} C_2 a^{k-1} b^2+$
$ \ldots+{ }^{k+1} C_k a b^k+{ }^{k+1} C_{k+1} b^{k+1} $
$ \left(\text { using }{ }^{k+1} C_0=1,{ }^k C_r+{ }^k C_{r-1}={ }^{k+1} C_r \text { and }{ }^k C_k=1={ }^{k+1} C_{k+1}\right) $
Thus, it has been proved that $P(k+1)$ is true whenever $P(k)$ is true. Therefore, by principle of mathematical induction, $P(n)$ is true for every positive integer $n$.
Hence,
$(a+b)^n={ }^n C_0 a^n+{ }^n C_1 a^{n-1} b+\ldots+{ }^n C_{n-1} a b^{n-1}+{ }^n C_n b^n$