( 0 1 6 ) + n = 1 ∑ 1 6 ( 2 n 2 + 7 n − 1 ) ( n 1 6 ) = ?
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Could you please explain the third step and the latter ?
Okay got it! You used ( r n ) = r n ( r − 1 n − 1 ) Ryt ?
Its a bad question!! So much calculative -_-
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Relevant wiki: Properties of Binomial Coefficients
( 0 1 6 ) + n = 1 ∑ 1 6 ( 2 n 2 + 7 n − 1 ) ( n 1 6 )
= 1 + n = 1 ∑ 1 6 ( 2 n ( n − 1 ) + 9 n − 1 ) ( n 1 6 )
= 1 + n = 1 ∑ 1 6 ( 3 2 ( n − 1 ) + 1 4 4 ) ( n − 1 1 5 ) − ( 2 1 6 − 1 )
= 2 − 2 1 6 + n = 1 ∑ 1 6 ( 3 2 ( n − 1 ) ) ( n − 1 1 5 ) + 1 4 4 × 2 1 5
= 2 − 2 1 6 + 1 4 4 × 2 1 5 + n = 2 ∑ 1 6 ( 3 2 ( n − 1 ) ) ( n − 1 1 5 )
= 2 + 1 4 2 × 2 1 5 + n = 2 ∑ 1 6 4 8 0 ( n − 2 1 4 )
= 2 + 1 4 2 × 2 1 5 + 4 8 0 × 2 1 4
= 2 + 7 6 4 × 2 1 4
= 1 2 5 1 7 3 7 8