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Question: Which of the following option(s) is/are correct?...

Which of the following option(s) is/are correct?

A

16C0+16C1+16C2+...+16C16=216{^{16}}C_0 + {^{16}}C_1 + {^{16}}C_2 + ... + {^{16}}C_{16} = 2^{16}

B

16C0+16C2+16C4+...+16C16=215{^{16}}C_0 + {^{16}}C_2 + {^{16}}C_4 + ... + {^{16}}C_{16} = 2^{15}

C

16C1+16C3+16C5+...+16C15=215{^{16}}C_1 + {^{16}}C_3 + {^{16}}C_5 + ... + {^{16}}C_{15} = 2^{15}

D

16C0−16C1+16C2−16C3+...+16C16=215{^{16}}C_0 - {^{16}}C_1 + {^{16}}C_2 - {^{16}}C_3 + ... + {^{16}}C_{16} = 2^{15}

Answer

A, B, C

Explanation

Solution

The binomial expansion of (1+x)n(1+x)^n is ∑r=0nnCrxr\sum_{r=0}^n {^n}C_r x^r. Setting x=1x=1 gives ∑r=0nnCr=2n\sum_{r=0}^n {^n}C_r = 2^n, proving option A for n=16n=16. Setting x=−1x=-1 gives ∑r=0n(−1)rnCr=0\sum_{r=0}^n (-1)^r {^n}C_r = 0 (for n≥1n \ge 1), disproving option D. Adding the expansions for x=1x=1 and x=−1x=-1 gives 2∑r evennCr=2n2 \sum_{r \text{ even}} {^n}C_r = 2^n, so ∑r evennCr=2n−1\sum_{r \text{ even}} {^n}C_r = 2^{n-1}, proving option B. Subtracting the x=−1x=-1 expansion from the x=1x=1 expansion gives 2∑r oddnCr=2n2 \sum_{r \text{ odd}} {^n}C_r = 2^n, so ∑r oddnCr=2n−1\sum_{r \text{ odd}} {^n}C_r = 2^{n-1}, proving option C.