Question

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

• A. ${}^n{C}_1{a}^n+{}^n{C}_2{a}^{n-1}b+{}^n{C}_3{a}^{n-2}{b}^2+\dots \dots \dots +{}^n{C}_{n+1}{b}^n$
• B. ${}^n{C}_0{b}^n+{}^n{C}_1{b}^{n-3}a+{}^n{C}_2{b}^{n-5}{a}^2+\dots ..+{}^n{C}_n{a}^n$
• C. ${}^n{C}_0{a}^n+{}^n{C}_1{a}^{n-1}b+{}^n{C}_2{a}^{n-2}{b}^2+\dots \dots +{}^n{C}_{n-1}a{b}^{n-1}+{}^n{C}_n{b}^n$
• D. None of the above

Answer 1

Answer: (C)

The proof of this result is given below
Let the given statement be
$P(n):(a+b{)}^n={}^n{C}_0{a}^n+{}^n{C}_1{a}^{n-1}b+{}^n{C}_2{a}^{n-2}{b}^2+$
$\dots +{}^n{C}_{n-1}a\cdot b^{n-1}+{}^n{C}_n{b}^n$
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.,
$(a+b{)}^k={}^k{C}_0{a}^k+{}^k{C}_1{a}^{k-1}b+{}^k{C}_2{a}^{k-2}{b}^2+\dots +{}^k{C}_k{b}^k\dots \text{ (i) }$
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, $(a+b{)}^{k+1}=(a+b)(a+b{)}^k$
$=(a+b)({}^k{C}_{0}a{}^k+{}^k{C}_1{a}^{k-1}b+{}^k{C}_2{a}^{k-2}{b}^2+$
$\dots +{}^k{C}_{k-1}a{b}^{k-1}+{}^k{C}_k{b}^k)[\text{ from Eq. (i) }]$
$={}^k{C}_{0}a{}^{k+1}+{}^k{C}_1{a}^{k}b+{}^k{C}_{2}a{}^{k-1}{b}^2+\dots +{}^k{C}_{k-1}{a}^2{b}^{k-1}$
$+{}^k{C}_{k}a{b}^k+{}^k{C}_0{a}^{k}b+{}^k{C}_1{a}^{k-1}{b}^2+{}^k{C}_2{a}^{k-2}{b}^3+$
$\dots +{}^k{C}_{k-1}a{b}^k+{}^k{C}_k{b}^{k+1}\text{ (by actual multiplication) }$
$={}^k{C}_0{a}^{k+1}+\left({}^k{C}_1+{}^k{C}_0\right)a^{k}b+\left({}^k{C}_2+{}^k{C}_1\right)a^{k-1}{b}^2+$
$\dots +\left({}^k{C}_k+{}^k{C}_{k-1}\right)a{b}^k+{}^k{C}_k{b}^{k+1}\text{ (grouping like terms) }$
$={}^{k+1}{C}_0{a}^{k+1}+{}^{k+1}{C}_1{a}^{k}b+{}^{k+1}{C}_2{a}^{k-1}{b}^2+$
$\dots +{}^{k+1}{C}_{k}a{b}^k+{}^{k+1}{C}_{k+1}{b}^{k+1}$
$\left(\text{ using }{}^{k+1}{C}_0=1,{}^k{C}_r+{}^k{C}_{r-1}={}^{k+1}{C}_r\text{ and }{}^k{C}_k=1={}^{k+1}{C}_{k+1}\right)$
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,
$(a+b{)}^n={}^n{C}_0{a}^n+{}^n{C}_1{a}^{n-1}b+\dots +{}^n{C}_{n-1}a{b}^{n-1}+{}^n{C}_n{b}^n$