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.