
Number of dissimilar terms in the expansion of $ {\left( {{x_1} + {x_2} + .... + {x_n}} \right)^3} $
This question has multiple correct options
(A) $ \dfrac{{{n^2}{{(n + 1)}^2}}}{4} $
(B) $ \dfrac{{n(n + 1)(n + 2)}}{6} $
(C) \[{}^{n + 1}{C_2} + {}^n{C_2}\]
(D) $ \dfrac{{{n^3} + 3{n^2}}}{4} $
Answer
576.3k+ views
Hint: Use the formula for the number of terms of a multinomial expansion. Put 3 in it to get the answer in terms of combination. Then expand the combination that you get and see for which options your solution matches.
Complete step-by-step answer:
The equation given in the question is
$ {\left( {{x_1} + {x_2} + .... + {x_n}} \right)^3} $
We have the formula for the number of terms of a multinomial expansion $ {\left( {{x_1} + {x_2} + .... + {x_n}} \right)^r} $ as
$ N = {}^{r + n - 1}{C_{n - 1}} $
Where,
$ N $ is the number of terms
$ r $ is the degree of the expansion
$ n $ is the number of variables in the expansion
For this question $ r = 3 $
Therefore, number of dissimilar terms are
$ N = {}^{3 + n - 1}{C_{n - 1}} $
$ \Rightarrow N = {}^{n + 2}{C_{n - 1}} $
Now, we know that the combination $ {}^n{C_r} $ can be expanded as
$ {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} $
Using the above formula, we can write
$ N = \dfrac{{(n + 2)!}}{{(n - 1)![(n + 2) - (n - 1)]!}} $
$ \Rightarrow N = \dfrac{{(n + 2)!}}{{(n - 1)![n + 2 - n + 1]!}} $
$ \Rightarrow N = \dfrac{{(n + 2)!}}{{(n - 1)! \times 3!}} $
Now we know that, factorial can be expanded as
$ n! = n(n - 1)(n - 2)....3.2.1 $ and $ n! = n(n - 1)! $
Using these two properties, we can write
$ N = \dfrac{{(n + 2)(n + 1)(n)(n - 1)!}}{{(n - 1)! \times 3!}} $
By cancelling the common terms, we get
$ N = \dfrac{{(n + 2)(n + 1)(n)}}{{3!}} $
Rearranging it we can write
$ N = \dfrac{{n(n + 2)(n + 1)}}{{3 \times 2 \times 1}} $
$ \Rightarrow N = \dfrac{{n(n + 2)(n + 1)}}{6} $
Thus, option (B) is correct.
Clearly, option (C) \[{}^{n + 1}{C_2} + {}^n{C_2}\] is incorrect as it has a constant term $ = 2 $ at the base of the combination. Whereas our solution has $ n - 1 $ .
Option (A) and option (D) are incorrect because they have 4 in the denominator but our solution has 6 in the denominator. You can also check that the numerators of both the options do not match with the numerator of the solution that we got.
Therefore, from the above explanation, the correct answer is, option (B) $ \dfrac{{n(n + 2)(n + 1)}}{6} $
So, the correct answer is “Option C”.
Note: In this question, it is given that there can be multiple answers. In such cases if you get only one correct answer then don’t get confused or don’t think that you have done something wrong. Just find out the reasons why you think other answers are incorrect and be confident about it. In this question, we have used the formula of number of terms. But the question was about a number of dissimilar terms. Still it didn’t make any difference because all the terms given in the multinomial are different.
Complete step-by-step answer:
The equation given in the question is
$ {\left( {{x_1} + {x_2} + .... + {x_n}} \right)^3} $
We have the formula for the number of terms of a multinomial expansion $ {\left( {{x_1} + {x_2} + .... + {x_n}} \right)^r} $ as
$ N = {}^{r + n - 1}{C_{n - 1}} $
Where,
$ N $ is the number of terms
$ r $ is the degree of the expansion
$ n $ is the number of variables in the expansion
For this question $ r = 3 $
Therefore, number of dissimilar terms are
$ N = {}^{3 + n - 1}{C_{n - 1}} $
$ \Rightarrow N = {}^{n + 2}{C_{n - 1}} $
Now, we know that the combination $ {}^n{C_r} $ can be expanded as
$ {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} $
Using the above formula, we can write
$ N = \dfrac{{(n + 2)!}}{{(n - 1)![(n + 2) - (n - 1)]!}} $
$ \Rightarrow N = \dfrac{{(n + 2)!}}{{(n - 1)![n + 2 - n + 1]!}} $
$ \Rightarrow N = \dfrac{{(n + 2)!}}{{(n - 1)! \times 3!}} $
Now we know that, factorial can be expanded as
$ n! = n(n - 1)(n - 2)....3.2.1 $ and $ n! = n(n - 1)! $
Using these two properties, we can write
$ N = \dfrac{{(n + 2)(n + 1)(n)(n - 1)!}}{{(n - 1)! \times 3!}} $
By cancelling the common terms, we get
$ N = \dfrac{{(n + 2)(n + 1)(n)}}{{3!}} $
Rearranging it we can write
$ N = \dfrac{{n(n + 2)(n + 1)}}{{3 \times 2 \times 1}} $
$ \Rightarrow N = \dfrac{{n(n + 2)(n + 1)}}{6} $
Thus, option (B) is correct.
Clearly, option (C) \[{}^{n + 1}{C_2} + {}^n{C_2}\] is incorrect as it has a constant term $ = 2 $ at the base of the combination. Whereas our solution has $ n - 1 $ .
Option (A) and option (D) are incorrect because they have 4 in the denominator but our solution has 6 in the denominator. You can also check that the numerators of both the options do not match with the numerator of the solution that we got.
Therefore, from the above explanation, the correct answer is, option (B) $ \dfrac{{n(n + 2)(n + 1)}}{6} $
So, the correct answer is “Option C”.
Note: In this question, it is given that there can be multiple answers. In such cases if you get only one correct answer then don’t get confused or don’t think that you have done something wrong. Just find out the reasons why you think other answers are incorrect and be confident about it. In this question, we have used the formula of number of terms. But the question was about a number of dissimilar terms. Still it didn’t make any difference because all the terms given in the multinomial are different.
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