Page 152 - Fixtures

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n , .

∑C (n,k)
k=0

.n

n

∑C (n,k) = 2n

k=0

2n = (1 + 1)n :

2n = C (n, 0)1n × 10 + C(n,1)× 1n−1 ×11 + + C (n,n)× 10 ×1n

n

= C(n, 0) + C(n,1) + + C(n,n) = ∑C(n,k)
k =0

[Column-Sum of Pascal Triangle]

, ()

60

1 + 3 + 6 + 10 + 15 = 35

.

:n

n0 r

.r +1 n +1

. n = C (n + 1, r + 1)

∑C (k,r)
k=0