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The value of _{j=1}^{n}(^{n+1}C_{j}-^{n}C_{j})is equal to

Question

The value of $j = 1 \sum n (^{n + 1} C_{j} -^{n} C_{j})$ is equal to

2^{n}

2^{n} + 1

3 \cdot 2^{n}

2^{n} - 1

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Jack Discussed

The value of

j = 1 \sum n (^{n + 1} C_{j} -^{n} C_{j})

is equal to

15 mins ago

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15 mins ago

Text solutionVerified

Simplifying, we get

(^{n + 1} C_{1} +^{n + 1} C_{2} + ...^{n + 1} C_{n}) - (^{n} C_{1} +^{n} C_{2} + ...^{n} C_{n})

= (^{n + 1} C_{1} +^{n + 1} C_{2} + ...^{n + 1} C_{n} +^{n + 1} C_{n + 1} -^{n + 1} C_{n + 1}) - 2^{n}

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

= 2^{n} (2 - 1) - 1

= 2^{n} - 1

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Question Text	The value of $j = 1 \sum n (^{n + 1} C_{j} -^{n} C_{j})$ is equal to
Answer Type	Text solution:1
Upvotes	150