Question:
Expand $(1 - x + x^2)^4$.
Expand $(1 - x + x^2)^4$.
Updated On: Jul 6, 2022
- $ 1-4 x + 10x^{2}- 16x^{3}+ 19x^{4}- 16x^{2}+ 10x^{6}-4x^{7} + x^{8}$
- $1-4 x + 10x^{2}- 16x^{3}+ x^{4}+ 16x^{5}+ 10x^{6}+4x^{7} - x^{8}$
- $1-4 x -10x^{2}+ 16x^{3}-19 x^{4}+ 16x^{5}- 10x^{6}+4x^{7} + x^{8}$
- $1-4 x -10x^{2}- 16x^{3}+ 19x^{4}- 16x^{5} + x^{8}$
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The Correct Option is A
Solution and Explanation
Put $1 - x = y$, then
$(1 - x + x^2)^4 = (y + x^2)^44$
$=\,^{4}C_{0}\,y^{4}+\,^{4}C_{1}\,y^{3}\left(x^{2}\right)^{1}+\,^{4}C_{2}\,y^{2}\left(x^{2}\right)^{2}+\,^{4}C_{3}\,y\left(x^{2}\right)^{3}$
$+\,^{4}C_{4}\left(x^{2}\right)^{4}$
$= y^{4} + 4y^{3}\, x^{2} + 6y^{2}\, x^{4} + 4y\, x^{6} + x^{8}$
$= \left(1 - x\right)^{4} + 4x^{2}\left(1 - x\right)^{3} + 6x^{4}\left(1 - x\right)^{2} + 4x^{6}\left(1 - x\right) + x^{8}$
$= 1-4 x + 10x^{2}- 16x^{3}+ 19x^{4}- 16x^{2}+ 10x^{6}-4x^{7} + x^{8}$
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Concepts Used:
Binomial Theorem
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
Properties of Binomial Theorem
- The number of coefficients in the binomial expansion of (x + y)n is equal to (n + 1).
- There are (n+1) terms in the expansion of (x+y)n.
- The first and the last terms are xn and yn respectively.
- From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
- The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.