Question

$\left( {n + 2} \right){}^n{C_0}{2^{n + 1}} - \left( {n + 1} \right){}^n{C_1}{2^n} + n{}^n{C_2}{2^{n - 1}} - ...$ is equal to
(A) 4
(B) $4n$
(C) $4\left( {n + 1} \right)$
(D) $2\left( {n + 2} \right)$

Answer 1

Hint : This question is based on the special cases of the Binomial Theorem. In this question the expanded form of the binomial expression is given and we have to find its value and write it in the simplest form possible. So, to simplify this expansion we can assume the smallest value of $n$ and substitute into the expression, then the value obtained would be also in the smallest form possible.

Complete step-by-step answer :
The expression given is –
$\left( {n + 2} \right){}^n{C_0}{2^{n + 1}} - \left( {n + 1} \right){}^n{C_1}{2^n} + n{}^n{C_2}{2^{n - 1}} - ...$
To simplify this expression, we have to assume a smallest value for the variable n.
Let us assume, the smallest value of n is $n = 1$ , then substituting this value of n into the expression we get,
$\Rightarrow \left( {n + 2} \right){}^n{C_0}{2^{n + 1}} - \left( {n + 1} \right){}^n{C_1}{2^n} + n{}^n{C_2}{2^{n - 1}} - ... = \left( {1 + 2} \right){}^1{C_0}{2^{1 + 1}} - \left( {1 + 1} \right){}^1{C_1}{2^1}$
By substituting $n = 1$ into the expression the terms after the second term of the expression are neglected because according to the binomial expansion the number of terms in the expansion for $n = 1$ should be $1 + 1 = 2$ . So, we consider only the first two terms of the expansion and rest are neglected.

$$\Rightarrow \left( {n + 2} \right){}^n{C_0}{2^{n + 1}} - \left( {n + 1} \right){}^n{C_1}{2^n} = \left( 3 \right)\dfrac{{1!}}{{\left( {1 - 0} \right)! \times 0!}}{2^2} - \left( 2 \right)\dfrac{{1!}}{{\left( {1 - 1} \right)! \times 1!}}{2^1}\\\ \Rightarrow \left( {n + 2} \right){}^n{C_0}{2^{n + 1}} - \left( {n + 1} \right){}^n{C_1}{2^n} = \left( {3 \times 4} \right)1 - \left( {2 \times 2} \right)1\\\ \Rightarrow \left( {n + 2} \right){}^n{C_0}{2^{n + 1}} - \left( {n + 1} \right){}^n{C_1}{2^n} = 12 - 4$$

So, for $n = 1$ the value of the expression is $12 - 4 = 8$
Now we can write the answer $8$ in the following way –
$\Rightarrow \left( {n + 2} \right){}^n{C_0}{2^{n + 1}} - \left( {n + 1} \right){}^n{C_1}{2^n} = 8\\\ \Rightarrow \left( {n + 2} \right){}^n{C_0}{2^{n + 1}} - \left( {n + 1} \right){}^n{C_1}{2^n} = 4 \times 2\\\ \Rightarrow \left( {n + 2} \right){}^n{C_0}{2^{n + 1}} - \left( {n + 1} \right){}^n{C_1}{2^n} = 4 \times \left( {1 + 1} \right)$
And we know the value of $n$ is $n = 1$ so substituting $n$ in place of 1, we get,
$\Rightarrow \left( {n + 2} \right){}^n{C_0}{2^{n + 1}} - \left( {n + 1} \right){}^n{C_1}{2^n} = 4 \times \left( {n + 1} \right)\\\ \Rightarrow \left( {n + 2} \right){}^n{C_0}{2^{n + 1}} - \left( {n + 1} \right){}^n{C_1}{2^n} = 4\left( {n + 1} \right)$
This is the required value of the expression in the simplest terms containing $n$ .
Therefore, the value of the expression $\left( {n + 2} \right){}^n{C_0}{2^{n + 1}} - \left( {n + 1} \right){}^n{C_1}{2^n} + n{}^n{C_2}{2^{n - 1}} - ...$ is equal to $4\left( {n + 1} \right)$ and the correct option is –
(C) $4\left( {n + 1} \right)$

So, the correct answer is “Option C”.

Note : The number of terms in the expansion of the binomial is defined by the value of the exponent of the binomial. For example, if the value of the exponent of a binomial is 2 then the number of terms in the binomial expansion would be $2 + 1 = 3.$
Also, for the binomial ${\left( {1 + x} \right)^2}$ the value of the exponent is 2 so the number of terms in the binomial expansion would be $2 + 1 = 3$ .