What Pascal Identity Should I use?
Let b k , n be the k th element of the n th row of Pascal's triangle, with row 1 being { 1 } , row 2 being { 1 , 1 } , etc. Also, the 1 st element of each row is 1 .
Find the value of
( b 3 , 2 0 1 5 ) 2 − i = 1 ∑ 2 0 1 3 i 3 .
Details and Assumptions:
- For example, b 1 , 1 = 1 , b 2 , 3 = 2 , and b 5 , 6 = 5 .
Image Credit: Wikipedia Pascal Triangle
The answer is 0.
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2 solutions
n − 1 C k − 1 = b k , n
Yes, but the C notation is obsolete. We use \binom {n-1}{k-1} ( k − 1 n − 1 ) or \dbinom {n-1}{k-1} ( k − 1 n − 1 ) ("d" for display).
It's actually very simple if you can see the identities involved. All of the following can be proven by induction if you wish.
Note that b 3 , n + 2 = T n , n ≥ 0 where T n is the n th triangular number. Therefore b 3 , 2 0 1 5 = T 2 0 1 3 .
Also, ∑ i = 1 n i 3 = T n 2 . Therefore i = 1 ∑ 2 0 1 3 i 3 = T 2 0 1 3 2 . T 2 0 1 3 2 − T 2 0 1 3 2 = 0
You could also use b k , n = ( k − 1 n − 1 )
Pascal's triangle gives the binomial coefficients. Therefore,
b k , n = ( n − 1 k − 1 )
⇒ b 3 , 2 0 1 5 − i = 1 ∑ 2 0 1 3 i 3 ⇒ ( 2 0 1 4 2 ) 2 − ( 2 2 0 1 3 × 2 0 1 4 ) 2 ⇒ ( 2 2 0 1 3 × 2 0 1 4 ) 2 − ( 2 2 0 1 3 × 2 0 1 4 ) 2 = 0