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Difficulty 5.7 AIME, harder Prove it
Example 1: Prove that ∑ k = 1 50 C 1948 + k k = C 1999 1949 − 1 \sum_{k=1}^{50} \mathrm{C}_{1948+k}^{k}=\mathrm{C}_{1999}^{1949}-1 ∑ k = 1 50 C 1948 + k k = C 1999 1949 − 1
Analysis: Observing the characteristics of the equation, it is easy to think of using the properties of binomial coefficients C n k = C n n − k \mathrm{C}_{n}^{k}=\mathrm{C}_{n}^{n-k} C n k = C n n − k C n k − 1 + C n k = C n + 1 k \mathrm{C}_{n}^{k-1}+\mathrm{C}_{n}^{k}=\mathrm{C}_{n+1}^{k} C n k − 1 + C n k = C n + 1 k
Solution Prove: ∑ k = 1 50 C 1948 + k k = ∑ k = 1 50 C 1948 + k 1948 = C 1999 1949 + ∑ k = 1 50 C 1948 + k 1948 − 1 = C 1999 1949 − 1. \begin{aligned} & \sum_{k=1}^{50} \mathrm{C}_{1948+k}^{k}=\sum_{k=1}^{50} \mathrm{C}_{1948+k}^{1948} \\ & =\mathrm{C}_{1999}^{1949}+\sum_{k=1}^{50} \mathrm{C}_{1948+k}^{1948}-1 \\ & =\mathrm{C}_{1999}^{1949}-1 .\end{aligned} k = 1 ∑ 50 C 1948 + k k = k = 1 ∑ 50 C 1948 + k 1948 = C 1999 1949 + k = 1 ∑ 50 C 1948 + k 1948 − 1 = C 1999 1949 − 1.
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