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If the number of terms in the expansion of ${(1 + 5 x + 10 x^{2} + 10 x^{3} + 5 x^{4} + x^{5})}^{20}$ is $m,$ then unit’s place of $2^{m}$ is

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Correct option is A

1+5x+10x2+10x3+5x4+x520=(1+x)520=(1+x)100m = number of terms = 101,We have2m=2101=2520=(30+2)20(2)=10m1+220Let k = Unit's digit of 2m is= units digit of 221Now, 221=254(2)=(30+2)4(2)∴ k = unit’s digit of 25=2.

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If the number of terms in the expansion of 1+5x+10x2+10x3+5x4+x520 is m, then unit’s place of 2m is

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If the number of terms in the expansion of ${(1 + 5 x + 10 x^{2} + 10 x^{3} + 5 x^{4} + x^{5})}^{20}$ is $m,$ then unit’s place of $2^{m}$ is