+289 Source: Mu Alpha Theta
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Evaluate:
\(\binom{15}{0} + 3\binom{15}{1}+....+ 31\binom{15}{15}\)
Your answer should be in the form of 2^n.
Hint:https://web2.0calc.com/questions/double-counting
listfor(n, 0, 15, ((2*n + 1)*(15 nCr n))==(1, 45, 525, 3185, 12285, 33033, 65065, 96525, 109395, 95095, 63063, 31395, 11375, 2835, 435, 31)>>>Sum==524,288
Interesting Problem!!!!!!
\(\text{ Let } S=\)\(15C0+3(15C1)+5(15C2)+7(15C3)+...+31(15C15)\)
Then, \(S=(15C15)+3(15C14)+...+31(15C0)\) (Using the property that: \(nCk = nC(n-k)\) ).
Adding the expressions above, to get:
\(2S=(15C0)(31+1)+(15C1)(3+29)+...+(15C15)(31+1)\)
Hence, \(2S=32(15C0+15C1+15C2+...+15C15)=32*2^{15}=2^5*2^{15}=2^{20}\)
Therefore, \(S=2^{19}\)
