If $C_{r}$ stands for ${^{n} C}_{r}$ then $(C_{0} + C_{1}) + (C_{1} + C_{2}) + . . . . (C_{n - 1} + C_{n})$ is equal to

2^{n} - 1

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2^{n + 1} + 1

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2^{n + 1} - 1

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2^{n + 1} - 2

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Solution

The correct option is D $2^{n + 1} - 2$
$(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x^{3} . . . . . . . . . . . + C_{n} x^{n}$

Put x =1

Using the relation of sum of coefficent
$(C_{0} + C_{1} +$ $C_{2} + C_{3} . . . . C_{n}) = 2^{n}$
According to the question ,

$(C_{0} + C_{1}) +$ $(C_{1} + C_{2}) +$ $(C_{2} + C_{3}) +$ $(C_{n - 1} + C_{n}) \to$

on simplifying
$(C_{0} + C_{1} +$ $C_{2} + C_{3} . . . . C_{n}) +$ $(C_{1} + C_{2} + C_{3} + . . . .$ $C_{n - 1}) \to$

Adding and subtracting $C_{0} a n d C_{n}$ to above equation

$(C_{0} + C_{1} +$ $C_{2} + C_{3} . . . . C_{n}) +$ $(C_{0} + C_{1} + C_{2} + C_{3} + . . . .$ $C_{n}) - C_{0} - C_{n} \to$

$= 2^{n} + 2^{n} - 1 - 1 = 2^{n + 1} - 2$

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