First slide

Binomial theorem for positive integral Index

Question

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ , then $C_{0} - (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) - (C_{0} + C_{1} + C_{2} + C_{3}) + \dots (- 1)^{n - 1} (C_{0} + C_{1} + \dots + C_{n - 1})$ is (where n is even integer and $C_{r} =^{n} C_{r}$ )

Moderate

A

a positive value
B

a negative value
C

divisible by 2^n-1
D

divisible by 2ⁿ

Solution

For n = 2m, the given expression is
$C_{0} - (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) - (C_{0} + C_{1} + C_{2} + C_{3}) + \dots (- 1)^{n - 1} (C_{0} + C_{1} + \dots + C_{n - 1})$
$= C_{0} - (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) - (C_{0} + C_{1} + C_{2} + C_{3}) + \dots - (C_{0} + C_{1} + \dots + C_{2 m - 1}) = - (C_{1} + C_{3} + C_{5} + \dots + C_{2 m - 1}) = - (C_{1} + C_{3} + C_{5} + \dots + C_{n - 1}) = - 2^{n - 1}$

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