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Question: Evaluate the series, $^{14}C_{14}$ $^{20}C_{10}$+$^{15}C_{14}$ $^{19}C_{10}$+$^{16}C_{14}$ $^{18}C_{...

Evaluate the series, 14C14^{14}C_{14} 20C10^{20}C_{10}+15C14^{15}C_{14} 19C10^{19}C_{10}+16C14^{16}C_{14} 18C10^{18}C_{10}.......24C14^{24}C_{14} 10C10^{10}C_{10}

A

34C24^{34}C_{24}

B

24C10^{24}C_{10}

C

35C10^{35}C_{10}

D

24C14^{24}C_{14}

Answer

35C10^{35}C_{10}

Explanation

Solution

The given series can be written as:

S=∑k=01014+kC14⋅20−kC10S = \sum_{k=0}^{10} {^{14+k}C_{14} \cdot ^{20-k}C_{10}}

Using the identity nCr=nCn−r^{n}C_r = ^{n}C_{n-r}, we can rewrite the terms:

14+kC14=14+kCk^{14+k}C_{14} = ^{14+k}C_k
20−kC10=20−kC10−k^{20-k}C_{10} = ^{20-k}C_{10-k}

The series becomes:

S=∑k=01014+kCk⋅20−kC10−kS = \sum_{k=0}^{10} {^{14+k}C_k \cdot ^{20-k}C_{10-k}}

This matches the form of the identity:

∑k=0n(r+kk)(s+n−kn−k)=(r+s+n+1n)\sum_{k=0}^n \binom{r+k}{k} \binom{s+n-k}{n-k} = \binom{r+s+n+1}{n}

with n=10n=10, r=14r=14, and s=10s=10. Applying the identity, we get:

S=(14+10+10+110)=(3510)=35C10S = \binom{14+10+10+1}{10} = \binom{35}{10} = {^{35}C_{10}}