Binomial Madness

Algebra Level 5

( 16 0 ) + n = 1 16 ( 2 n 2 + 7 n 1 ) ( 16 n ) = ? \binom{16}{0}+\sum_{n = 1}^{16}(2n^2 + 7n - 1)\binom{16}{n} = \, ?


The answer is 12517378.

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1 solution

展豪 張
Apr 10, 2016

Relevant wiki: Properties of Binomial Coefficients

( 16 0 ) + n = 1 16 ( 2 n 2 + 7 n 1 ) ( 16 n ) \;\;\;\;\displaystyle\binom{16}{0}+\sum_{n = 1}^{16}(2n^2 + 7n - 1)\binom{16}n
= 1 + n = 1 16 ( 2 n ( n 1 ) + 9 n 1 ) ( 16 n ) =\displaystyle 1+\sum_{n=1}^{16}(2n(n-1)+9n-1)\binom{16}n
= 1 + n = 1 16 ( 32 ( n 1 ) + 144 ) ( 15 n 1 ) ( 2 16 1 ) =\displaystyle 1+\sum_{n=1}^{16}(32(n-1)+144)\binom{15}{n-1}-(2^{16}-1)
= 2 2 16 + n = 1 16 ( 32 ( n 1 ) ) ( 15 n 1 ) + 144 × 2 15 =\displaystyle 2-2^{16}+\sum_{n=1}^{16}(32(n-1))\binom{15}{n-1}+144\times 2^{15}
= 2 2 16 + 144 × 2 15 + n = 2 16 ( 32 ( n 1 ) ) ( 15 n 1 ) =\displaystyle 2-2^{16}+144\times 2^{15}+\sum_{n=2}^{16}(32(n-1))\binom{15}{n-1}
= 2 + 142 × 2 15 + n = 2 16 480 ( 14 n 2 ) =\displaystyle 2+142\times 2^{15}+\sum_{n=2}^{16}480\binom{14}{n-2}
= 2 + 142 × 2 15 + 480 × 2 14 =2+142\times 2^{15}+480\times 2^{14}
= 2 + 764 × 2 14 =2+764\times 2^{14}
= 12517378 =12517378


Could you please explain the third step and the latter ?

Anurag Pandey - 4 years, 8 months ago

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I have used i ( n i ) = 2 n \sum_i \binom n i =2^n .

展豪 張 - 4 years, 8 months ago

Okay got it! You used ( n r ) = n r ( n 1 r 1 ) \binom{n}{r}= \frac{n}{r} \binom{n-1}{r-1} Ryt ?

Anurag Pandey - 4 years, 8 months ago

Its a bad question!! So much calculative -_-

Md Zuhair - 3 years, 5 months ago

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