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Prove that \(\displaystyle\sum_{r=0}^{n} \)  nCr.3r = 4n

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To prove:   \(\displaystyle\sum_{r=0}^{n} \)  nCr.3r = 4n

Formula used:   \(\displaystyle\sum_{r=0}^{n} \)  nCr.an-rbr = (a+b)n

Proof: In the above formula if we put a = 1 and b = 3, then we will get

  \(\displaystyle\sum_{r=0}^{n} \)  nCr.1n-r3r = (1+3)n

Therefore, 

  \(\displaystyle\sum_{r=0}^{n} \)  nCr.3r = 4n

Hence Proved.

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