The coefficient of $x^{1012}$ in the expansion of $(1 + x^n + x^{253})^{10}$, (where n $\le$ 22 is any positive integer), is :-
- 1
- ${^{10}C_4}$
- 4n
- ${^{253}C_4}$
The Correct Option is B
Solution and Explanation
Given expansion \((1+ x^n + x^{253})^{10}\)
Let \(x^{1012} = (1)^a (x^n)^b. (x^{253})^c\)
Here a, b, c, n are all +ve integers and a \(\le\) 10, b \(\le\) 10, c \(\le\) 4, n \(\le\) 22, a + b + c = 10
Now \(bn + 253c = 1012\) \(\Rightarrow \, bn = 253 (4 - c)\)
For c < 4 and n \(\le\) 22; b > 10,
which is not possible.
\(\therefore\) c = 4, b = 0, a =6
\(\therefore \, x^{1012} = (1)^6. (x^n)^0 . (x^{253})^{4}\)
Hence the coefficient of \(x^{1012}\) \(= \frac{10!}{6! 0! 4!}\) \(= {^{10}C_4}\)
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Concepts Used:
Binomial Expansion Formula
The binomial expansion formula involves binomial coefficients which are of the form
(n/k)(or) nCk and it is calculated using the formula, nCk =n! / [(n - k)! k!]. The binomial expansion formula is also known as the binomial theorem. Here are the binomial expansion formulas.
This binomial expansion formula gives the expansion of (x + y)n where 'n' is a natural number. The expansion of (x + y)n has (n + 1) terms. This formula says:
We have (x + y)n = nC0 xn + nC1 xn-1 . y + nC2 xn-2 . y2 + … + nCn yn
General Term = Tr+1 = nCr xn-r . yr
- General Term in (1 + x)n is nCr xr
- In the binomial expansion of (x + y)n , the rth term from end is (n – r + 2)th .