If r=0∑n{′nCr+nCr−1nCr−1}3=2425, then n is equal to

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Question

If $r = 0 \sum n {\frac{^{n} C _{r - 1}}{^{' n} C _{r} + ^{n} C _{r - 1}}}^{3} = \frac{25}{24}$ , then $n$ is equal to

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Alexander Discussed

r = 0 \sum n {\frac{^{n} C _{r - 1}}{^{' n} C _{r} + ^{n} C _{r - 1}}}^{3} = \frac{25}{24}

, then

n

is equal to

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Question Text

r = 0 \sum n {\frac{^{n} C _{r - 1}}{^{' n} C _{r} + ^{n} C _{r - 1}}}^{3} = \frac{25}{24}

, then

n

is equal to