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Question: Find the sum of \( 1{}^n{C_0} + 3{}^n{C_1} + 5{}^n{C_2} + ..... + (2n + 1){}^n{C_n} is ?\) A. \( {...

Find the sum of 1nC0+3nC1+5nC2+.....+(2n+1)nCnis? 1{}^n{C_0} + 3{}^n{C_1} + 5{}^n{C_2} + ..... + (2n + 1){}^n{C_n} is ?
A. 2n+1(n+1){2^{n + 1}}(n + 1)
B. (n+1)2n(n + 1){2^n}
C. n.2n+1n{.2^{n + 1}}
D. None of the above

Explanation

Solution

Hint : The binomial expansion or the binomial theorem describes the algebraic expansion of the powers of the binomial (binomial is the pair of two terms). Use formula (1+x)n=nC1x1+nC2x2+.....{(1 + x)^n} = {}^n{C_1}{x^1} + {}^n{C_2}{x^2} + ..... for binomial expansion and then find the sum of the terms given.

Complete step by step solution:
Now, take the expansion of the binomial expansion-
(1+x)n=nC0x0+nC1x1+nC2x2+.....+nCnxn{(1 + x)^n} = {}^n{C_0}{x^0} + {}^n{C_1}{x^1} + {}^n{C_2}{x^2} + ..... + {}^n{C_n}{x^n} …. (A)
Replace by in the above equation –
(1+x2)n=nC0+nC1x2+nC2x4+.....+nCnx2x{(1 + {x^2})^n} = {}^n{C_0} + {}^n{C_1}{x^2} + {}^n{C_2}{x^4} + ..... + {}^n{C_n}{x^{2x}} ….. (B)
Multiply the above equation with “x” on both the sides of the equation –
x(1+x2)n=nC0x+nC1x3+nC2x5+.....+nCnx2x+1x{(1 + {x^2})^n} = {}^n{C_0}x + {}^n{C_1}{x^3} + {}^n{C_2}{x^5} + ..... + {}^n{C_n}{x^{2x + 1}}
Differentiate the above expression –
(1+x2)n+2nx2(1+x2)n−1=nC0+3nC1x2+5nC2x4+.....+(2n+1)nCnx2x{(1 + {x^2})^n} + 2n{x^2}{(1 + {x^2})^{n - 1}} = {}^n{C_0} + 3{}^n{C_1}{x^2} + 5{}^n{C_2}{x^4} + ..... + (2n + 1){}^n{C_n}{x^{2x}}
Place x=1x = 1 in the above equation –
(1+1)n+2n(1)2(1+(1)2)n−1=nC0+3nC1(1)2+5nC2(1)4+.....+(2n+1)nCn(1)2(1){(1 + 1)^n} + 2n{(1)^2}{(1 + {(1)^2})^{n - 1}} = {}^n{C_0} + 3{}^n{C_1}{(1)^2} + 5{}^n{C_2}{(1)^4} + ..... + (2n + 1){}^n{C_n}{(1)^{2(1)}}
Simplify the above equation –
2n+2n(2)n−1=nC0+3nC1+5nC2+.....+(2n+1)nCn{2^n} + 2n{(2)^{n - 1}} = {}^n{C_0} + 3{}^n{C_1} + 5{}^n{C_2} + ..... + (2n + 1){}^n{C_n}
By using the laws of power and exponent in the above equation and simplify. When bases are the same and multiplicative then the powers are added.
2n+2nn=nC0+3nC1+5nC2+.....+(2n+1)nCn{2^n} + {2^n}n = {}^n{C_0} + 3{}^n{C_1} + 5{}^n{C_2} + ..... + (2n + 1){}^n{C_n}
Take common multiple common on the left hand side of the equation –
2n(1+n)=nC0+3nC1+5nC2+.....+(2n+1)nCn{2^n}(1 + n) = {}^n{C_0} + 3{}^n{C_1} + 5{}^n{C_2} + ..... + (2n + 1){}^n{C_n}
The above equation can be re-written as-
nC0+3nC1+5nC2+.....+(2n+1)nCn=2n(n+1){}^n{C_0} + 3{}^n{C_1} + 5{}^n{C_2} + ..... + (2n + 1){}^n{C_n} = {2^n}(n + 1)
Hence, from the given multiple choices, the option B is the correct answer.
So, the correct answer is “Option B”.

Note : Know the correct formula for the expansion of the binomials. Follow the laws of the power and exponent to simplify the expression. When bases are the same powers are added in case of terms in multiplication.
Remember the seven basic rules of the exponent or the laws of exponents to solve these types of questions. Make sure to go through the below mentioned rules, it describes how to solve different types of exponents problems and how to add, subtract, multiply and divide the exponents.
Product of powers rule
Quotient of powers rule
Power of a power rule
Power of a product rule
Power of a quotient rule
Zero power rule
Negative exponent rule