The value of -j-1nn-1Cj-nCjis equal to

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Standard XII

Mathematics

Sum of Coefficients of All Terms

The value of ...

Question

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

2^{n}

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2^{n} + 1

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3 \cdot 2^{n}

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2^{n} - 1

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Solution

The correct option is C $2^{n} - 1$
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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The value of n∑j=1(n+1Cj−nCj)is equal to

The correct option is C 2n−1Simplifying, we get (n+1C1+n+1C2+...n+1Cn)−(nC1+nC2+...nCn)=(n+1C1+n+1C2+...n+1Cn+n+1Cn+1−n+1Cn+1)−2n=2n+1−2n−n+1Cn+1=2n(2−1)−1=2n−1

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