Question

The number of terms in the expansion ${\left( {a + b + c} \right)^{12}}$ is
A) 90
B) 91
C) 81
D) 80

Answer 1

Here, we will use the binomial theorem to first expand the given polynomial. After expanding the given polynomial, we will find each term in its expansion. This means that the number of terms in each particular term keeps increasing by 1. Then, we will use the formula of sum of $n$ terms in an AP, to find the required number of terms in the expansion of the given polynomial.

Formula Used:
We will use the following formulas:
${\left( {a + b} \right)^n} = {}^nC{_0}{a^n}{b^0} + {}^nC{_1}{a^{n - 1}}{b^1} + {}^nC{_2}{a^{n - 2}}{b^2} + ......{}^nC{_n}{a^0}{b^n}$
Sum of $n$terms$= \dfrac{{n\left( {n + 1} \right)}}{2}$

Complete step by step solution:
In order to find the expansion of ${\left( {a + b + c} \right)^{12}}$, we will use binomial theorem.
According to binomial theorem, it is possible to write ${\left( {a + b} \right)^n}$ such that:
${\left( {a + b} \right)^n} = {}^nC{_0}{a^n}{b^0} + {}^nC{_1}{a^{n - 1}}{b^1} + {}^nC{_2}{a^{n - 2}}{b^2} + ......{}^nC{_n}{a^0}{b^n}$
Hence, using this formula, by substituting $a = a$, $b = \left( {b + c} \right)$ and $n = 12$, we get,
${\left( {a + \left( {b + c} \right)} \right)^{12}} = {}^{12}C{_0}{a^{12}}{\left( {b + c} \right)^0} + {}^{12}C{_1}{a^{11}}{\left( {b + c} \right)^1} + {}^{12}C{_2}{a^{10}}{\left( {b + c} \right)^2} + ......{}^{12}C{_{12}}{a^0}{\left( {b + c} \right)^{12}}$
Simplifying the expression, we get
$\Rightarrow {\left( {a + \left( {b + c} \right)} \right)^{12}} = {a^{12}} + 12{a^{11}}\left( {b + c} \right) + {}^{12}C{_2}{a^{10}}{\left( {b + c} \right)^2} + ......{\left( {b + c} \right)^{12}}$
Now, if we look at the terms present in this expansion, we will notice that after the expansion, the number of terms in each term will be:
The first term, ${a^{12}} = 1$ term
The second term, $12{a^{11}}\left( {b + c} \right) = 2$ terms
The third term, ${}^{12}{C_2}{a^{10}}{\left( {b + c} \right)^2} = 3$terms
And so on…
Also, we will find the number of terms in the last term.
The last term, ${\left( {b + c} \right)^{12}} = \left( {12 + 1} \right) = 13$ terms
Hence, number of terms $= 1 + 2 + 3 + .....12 + 13$
Now, we know that sum of $n$ terms $= \dfrac{{n\left( {n + 1} \right)}}{2}$
But, here, there are $\left( {n + 1} \right)$ terms,
Hence, we will write the formula as: $\dfrac{{\left( {n + 1} \right)\left( {n + 2} \right)}}{2}$
Now, substituting the given value, $n = 12$, we get,
$\Rightarrow$ Number of terms $= \dfrac{{\left( {12 + 1} \right)\left( {12 + 2} \right)}}{2}$
Adding the terms in the numerator, we get
$\Rightarrow$ Number of terms $= \dfrac{{13 \times 14}}{2}$
Dividing the numerator by 2, we get
$\Rightarrow$ Number of terms $= 13 \times 7 = 91$ terms
Therefore, the number of terms in the expansion ${\left( {a + b + c} \right)^{12}}$ is 91

Hence, option B is the correct answer.

Note:
In algebra, binomial theorem describes the algebraic expansion of the powers of a binomial. Also, we have used the formula of sum of $n$ terms in an AP. While using this formula, we should keep in mind that in this question, there are $\left( {n + 1} \right)$ terms and hence, substitute the values accordingly. If we forget to do this, then, our whole answer will become wrong. Also, if we forget this formula, then, we can use the general formula of sum of $n$ terms of an AP, i.e. ${S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$.