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B
3n
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C
∑nr=0(−1)rCr4r
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D
∑nr=0nCr2r
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Solution
The correct options are B3n C∑nr=0(−1)rCr4r D∑nr=0nCr2r k=n∑k=0n−k∑r=0nCkn−kCr Now n−k∑0n−kCr=2n−r Therefore k=n∑k=0nCk2n−k =2nnC0+2n−1nC1+...20nCn =(2+1)n =3n This can re-written as =(4−1)n =r=n∑r=0(−1)r4n−r