Question:

The coefficient of x5x^5 in the expansion of (1+x2)5(1+x)4(1 + x^2)^5(1 + x)^4 is

Updated On: Mar 4, 2024
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The Correct Option is B

Solution and Explanation

We have,
(1+x2)5=5C0(x2)0+5C1(x2)1+5C2(x2)2\left(1+x^{2}\right)^{5}={ }^{5} C_{0}\left(x^{2}\right)^{0}+{ }^{5} C_{1}\left(x^{2}\right)^{1}+{ }^{5} C_{2}\left(x^{2}\right)^{2}
+5C3(x2)3+5C4(x2)4+5C5(x2)5+{ }^{5} C_{3}\left(x^{2}\right)^{3}+{ }^{5} C_{4}\left(x^{2}\right)^{4}+{ }^{5} C_{5}\left(x^{2}\right)^{5}
=1+5x2+10x4+10x6+5x8+x10= 1+5 x^{2}+10 x^{4}+10 x^{6}+5 x^{8}+x^{10}
(1+x)4=4C0x0+4C1x1+4C2x2+4C3x3+4C4x4(1+x)^{4}={ }^{4} C_{0} x^{0}+{ }^{4} C_{1} x^{1}+{ }^{4} C_{2} x^{2}+{ }^{4} C_{3} x^{3}+{ }^{4} C_{4} x^{4}
=1+4x+6x2+4x3+x4= 1+4 x+6 x^{2}+4 x^{3}+x^{4}
\therefore Coefficient of x5x^{5} in the product of
(1+x2)5(1+x)4\left(1+x^{2}\right)^{5}(1+x)^{4}
=(5x2)(4x3)+(10x4).(4x)=\left(5 x^{2}\right) \cdot\left(4 x^{3}\right)+\left(10 x^{4}\right).(4 x)
=20x5+40x5=20 x^{5}+40 x^{5}
=60x5=60\, x^{5}
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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.