Question

The total number of terms in the expansion of ${(x+y)}^{100}+{(x–y)}^{100}$after simplification is

• A. $51$
• B. $202$
• C. $100$
• D. $50$

Answer 1

The correct option is A

$51$

Explanation of the correct option.

If $x$ and $y$ are real number, then for all $n$ belongs to $N$,

Since, ${\left(x+y\right)}^n={}^{n}C_0{x}^n{y}^0+{}^{n}C_1{x}^{n-1}{y}^1+..............................{}^{n}C_n{x}^0{y}^n$

$$\begin{gathered} \Rightarrow {\left(x+y\right)}^{100}={}^{100}C_0{x}^{100}{y}^0+{}^{100}C_1{x}^{99}{y}^1+..............................+{}^{100}C_{100}{x}^0{y}^{100}....\left(1\right) \\ \Rightarrow {\left(x-y\right)}^{100}={}^{100}C_0{x}^{100}{y}^0-{}^{100}C_1{x}^{99}{y}^1+..............................+{}^{100}C_{100}{x}^0{y}^{100}....\left(2\right) \end{gathered}$$

Add equation $\left(1\right)$ and $\left(2\right)$,

$$\begin{gathered} {\left(x+y\right)}^{100}+{\left(x-y\right)}^{100}={}^{100}C_0{x}^{100}{y}^0+{}^{100}C_1{x}^{99}{y}^1+..........+{}^{100}C_{100}{x}^0{y}^{100}+{}^{100}C_0{x}^{100}{y}^0-{}^{100}C_1{x}^{99}{y}^1+........+{}^{100}C_{100}{x}^0{y}^{100} \\ \Rightarrow {\left(x+y\right)}^{100}+{(x-y)}^{100}=2\left[{}^{100}C_0{x}^{100}{y}^0+{}^{100}C_2{x}^{98}{y}^2+{}^{100}C_4{x}^{96}{y}^4+{}^{100}C_6{x}^{94}{y}^6+...............+{}^{100}C_{100}{x}^0{y}^{100}\right] \end{gathered}$$

We can see that, there are only even terms in the series starting from $0$ to $100$, thus there are $51$ terms.

Hence, $o\mathrm{ption}\left(\mathrm{A}\right)$ is the correct option, i.e. $51$.