First slide

Binomial theorem for positive integral Index

Question

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

Moderate

A

2
B

8
C

6
D

4

Solution

$\begin{array}{l} {(1 + 5 x + 10 x^{2} + 10 x^{3} + 5 x^{4} + x^{5})}^{20} \\ = {((1 + x)^{5})}^{20} = (1 + x)^{100} \end{array}$
$m$ = number of terms = 101,
We have
$\begin{array}{l} 2^{m} = 2^{101} = {(2^{5})}^{20} \\ = (30 + 2)^{20} (2) = (10 m_{1} + 2^{20}) \end{array}$
Let $k$ = Unit's digit of $2^{m}$ is
= units digit of $2^{21}$
Now, $2^{21} = {(2^{5})}^{4} (2) = (30 + 2)^{4} (2)$
$∴$ k = unit’s digit of $2^{5} = 2 .$

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