Product of binomial coefficients

Algebra Level 3

If P n P_{n} denotes the product of all the binomial coefficients in the expansion of ( 1 + x ) n (1+x)^n , then P n + 1 P n \frac{P_{n+1}}{P_{n}} equals

Clarification : Binomial Coefficients are ( n r ) {n \choose r} OR n C r ^{n}C_{r}

( n + 1 ) n + 1 ( n + 1 ) ! \frac{(n+1)^{n+1}}{(n+1)!} ( n + 1 ) n ! \frac{(n+1)}{n!} n n n ! \frac{n^{n}}{n!} ( n + 1 ) n ( n + 1 ) ! \frac{(n+1)^{n}}{(n+1)!} n n ( n + 1 ) ! \frac{n^{n}}{(n+1)!}

Ravi Dwivedi by Ravi Dwivedi , 24, India

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1 solution

Ravi Dwivedi
Oct 21, 2015

P n = n C 0 ⋅ n C 1 ⋅ n C 2 ⋅ ⋅ ⋅ n C n P_{n}= ^{n}C_{0} \cdot ^{n}C_{1} \cdot ^{n}C_{2} \cdot \cdot \cdot ^{n}C_{n}

P n + 1 = n + 1 C 0 ⋅ n + 1 C 1 ⋅ n + 1 C 2 ⋅ ⋅ ⋅ n + 1 C n + 1 P_{n+1}= ^{n+1}C_{0} \cdot ^{n+1}C_{1} \cdot ^{n+1}C_{2} \cdot \cdot \cdot ^{n+1}C_{n+1}

Now P n + 1 P n = n C 0 ⋅ n C 1 ⋅ n C 2 ⋅ ⋅ ⋅ n C n n + 1 C 0 ⋅ n + 1 C 1 ⋅ n + 1 C 2 ⋅ ⋅ ⋅ n + 1 C n + 1 \frac{P_{n+1}}{P_{n}} = \frac{^{n}C_{0} \cdot ^{n}C_{1} \cdot ^{n}C_{2} \cdot \cdot \cdot ^{n}C_{n}}{^{n+1}C_{0} \cdot ^{n+1}C_{1} \cdot ^{n+1}C_{2} \cdot \cdot \cdot ^{n+1}C_{n+1}}

⟹ P n + 1 P n = ( n + 1 C 1 n C 0 ) ( n + 1 C 2 n C 1 ) ( n + 1 C 3 n C 2 ) ⋅ ⋅ ⋅ ( n + 1 C n + 1 n C n ) \implies \frac{P_{n+1}}{P_{n}} =( \frac{^{n+1}C_{1}}{^{n}C_{0}})( \frac{^{n+1}C_{2}}{^{n}C_{1}})( \frac{^{n+1}C_{3}}{^{n}C_{2}}) \cdot \cdot \cdot ( \frac{^{n+1}C_{n+1}}{^{n}C_{n}})

⟹ P n + 1 P n = ( n + 1 1 ⋅ n C 0 n C 0 ) ( n + 1 2 ⋅ n C 1 n C 1 ) ( n + 1 3 ⋅ n C 2 n C 2 ) ⋅ ⋅ ⋅ ( n + 1 n + 1 ⋅ n C n n C n ) \implies \frac{P_{n+1}}{P_{n}} =( \frac{\frac{n+1}{1} \cdot^{n}C_{0}}{^{n}C_{0}})( \frac{\frac{n+1}{2} \cdot^{n}C_{1}}{^{n}C_{1}})( \frac{\frac{n+1}{3}\cdot ^{n}C_{2}}{^{n}C_{2}}) \cdot \cdot \cdot ( \frac{\frac{n+1}{n+1}\cdot^{n}C_{n}}{^{n}C_{n}})

⟹ P n + 1 P n = ( n + 1 1 ) ( n + 1 2 ) ( n + 1 3 ) ⋅ ⋅ ⋅ ( n + 1 n + 1 ) = ( n + 1 ) n + 1 ( n + 1 ) ! \implies \frac{P_{n+1}}{P_{n}} = (\frac{n+1}{1})(\frac{n+1}{2})(\frac{n+1}{3})\cdot \cdot \cdot (\frac{n+1}{n+1})= \boxed{\frac{(n+1)^{n+1}}{(n+1)!}}

10 Helpful 4 Interesting 3 Brilliant 1 Confused

Moderator note:

Good usage of ( n + 1 k + 1 ) = n + 1 k + 1 × ( n k ) { n+1 \choose k + 1 } = \frac{n+1}{k+1} \times { n \choose k } .

The first step need correction

Ahmed Nabih - 2 years, 9 months ago

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