The expression $(2 x - 1)^{5} + 5 (2 x - 1)^{4} + 10 (2 x - 1)^{2} + 5 (2 x - 1) + 1$ is equal to

16

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32

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16 x^{5}

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N o n e o f t h e s e

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Solution

The correct option is D $N o n e o f t h e s e$
${(2 x - 1)}^{5} + 5 {(2 x - 1)}^{4} + 10 {(2 x - 1)}^{2} + 5 (2 x - 1) + 1$
$^{5} C_{0} {(2 x)}^{5} -^{5} C_{1} {(2 x)}^{4} +^{5} C_{2} {(2 x)}^{3} -^{5} C_{3} {(2 x)}^{2} +^{5} C_{4} (2 x) -^{5} C_{5} [^{4} C_{0} {(2 x)}^{4} -^{4} C_{1} {(2 x)}^{3} +^{4} C_{2} {(2 x)}^{2} -^{4} C_{3} (2 x) +^{4} C_{4}]$
$+ 10 [^{2} C_{0} {(2 x)}^{2} -^{2} C_{1} (2 x) +^{2} C_{2}] + 10 x - 5 + 1$
$= 32 x^{5} - 80 x^{4} + 80 x^{3} - 40 x^{2} + 10 x - 1 + 5 (16 x^{4} - 32 x^{3} + 24 x^{2} - 8 x + 1) + 10 (4 x^{2} - 4 x + 1) + 10 x - 4$
$= 32 x^{5} - 80 x^{4} + 80 x^{3} - 40 x^{2} + 10 x - 1 + 80 x^{4} + 160 x^{3} + 120 x^{2} - 40 x + 5 + 40 x^{2} - 40 x + 10 + 10 x - 4$
$= 32 x^{5} - 80 x^{3} + 60 x + 120 x^{2}$

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