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Question: The value of \(C_{0} + 3C_{1} + 5C_{2} + ..... + (2n + 1)C_{n}\) is equal to...

The value of C0+3C1+5C2+.....+(2n+1)CnC_{0} + 3C_{1} + 5C_{2} + ..... + (2n + 1)C_{n} is equal to

A

2n2^{n}

B

2n+n.2n−12^{n} + n.2^{n - 1}

C

2n(n+1)2^{n}(n + 1)

D

None of these

Answer

2n(n+1)2^{n}(n + 1)

Explanation

Solution

We have C0+3C1+5C2+.....+(2n+1)CnC_{0} + 3C_{1} + 5C_{2} + ..... + (2n + 1)C_{n}

= ∑r=0n(2r+1)Cr=∑r=0n(2r+1)n⥂Cr\sum_{r = 0}^{n}{(2r + 1)C_{r}} = \sum_{r = 0}^{n}{(2r + 1)^{n} ⥂ C_{r}}= ∑r=0n2rnCr+∑r=0nnCr\sum_{r = 0}^{n}{2r^{n}C_{r}} + \sum_{r = 0}^{n}{nC_{r}}

= 2.∑r=1nr.nr.n−1⥂Cr−1+∑r=0nn⥂Cr2.\sum_{r = 1}^{n}{r.\frac{n}{r}.^{n - 1} ⥂ C_{r - 1} + \sum_{r = 0}^{n}{n ⥂ C_{r}}}

= 2n∑r=1nn−1Cr−1+∑r=0nnCr2n\sum_{r = 1}^{n}{n - 1C_{r - 1}} + \sum_{r = 0}^{n}{nC_{r}}=2n[(1+1)n−1]+[1+1]n2n\lbrack(1 + 1)^{n - 1}\rbrack + \lbrack 1 + 1\rbrack^{n}

= 2n.2n−1+2n=2n.[n+1]2n.2^{n - 1} + 2^{n} = 2^{n}.\lbrack n + 1\rbrack.

Trick: Put n=1n = 1 in given expansion 1⥂C0+3.1⥂C1=1+3=41 ⥂ C_{0} + 3.^{1} ⥂ C_{1} = 1 + 3 = 4.

Which is given by option (3) 2n.(n+1)=21(1+1)=42^{n}.(n + 1) = 2^{1}(1 + 1) = 4.