Binomial theorem for positive integral Index

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$If C_{0}, C_{1}, C_{2}, \dots . . C_{n} denote the coefficients of the binomial expansion (1 + x)^{n}, then the value of C_{1} + 3 C_{3} + 5 C_{5} + \dots + is$

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Solution

$\begin{array}{l} {(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x^{3} + .... + C_{n} x^{n} \\ Differentiate both sides \\ n {(1 + x)}^{n - 1} = C_{1} + 2 C_{2} x + 3 C_{3} x^{2} + ... + n C_{n} x^{n - 1} \\ put x = 1 \\ n (2^{n - 1}) = C_{1} + 2 C_{2} + 3 C_{3} + ... + n C_{n} \\ put x = - 1 \\ 0 = C_{1} - 2 C_{2} + 3 C_{3} + .... + {(- 1)}^{n - 1} n C_{n} \\ Adding the above two equations \\ 2 (C_{1} + 3 C_{3} + 5 C_{5} + ...) = n (2^{n - 1}) \\ C_{1} + 3 C_{3} + 5 C_{5} + ... = n (2^{n - 2}) \end{array}$

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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

If C0,C1,C2,…..Cn denote the coefficients of the binomial expansion (1+x)n , then the value of C1+3C3+5C5+…+ is

1+xn=C0+C1x+C2x2+C3x3+....+CnxnDifferentiate both sides n1+xn−1=C1+2C2x+3C3x2+...+nCnxn−1put x=1n2n−1=C1+2C2+3C3+...+nCnput x=−10=C1−2C2+3C3+....+−1n−1nCnAdding the above two equations2C1+3C3+5C5+...=n2n−1C1+3C3+5C5+...=n2n−2

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$If C_{0}, C_{1}, C_{2}, \dots . . C_{n} denote the coefficients of the binomial expansion (1 + x)^{n}, then the value of C_{1} + 3 C_{3} + 5 C_{5} + \dots + is$