The constant term in the expansion of left(1+ | vedclass

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(D) Given expression: $E = \left(1+\frac{1}{x}\right)^{20} \left(30x(1+x)^{29} + (1+x)^{30}\right)$ Simplify the expression: $E = \left(\frac{x+1}{x}\

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${}^{50}C_{20}+30 \cdot {}^{49}C_{19}$

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(D) Given expression: $E = \left(1+\frac{1}{x}\right)^{20} \left(30x(1+x)^{29} + (1+x)^{30}\right)$ Simplify the expression: $E = \left(\frac{x+1}{x}\right)^{20} (1+x)^{29} (30x + 1 + x) = \frac{(1+x)^{20}}{x^{20}} (1+x)^{29} (31x + 1) = \frac{(1+x)^{49}}{x^{20}} (31x + 1)$ $E = 31x \cdot \frac{(1+x)^{49}}{x^{20}} + \frac{(1+x)^{49}}{x^{20}} = 31 \frac{(1+x)^{49}}{x^{19}} + \frac{(1+x)^{49}}{x^{20}}$ The constant term is the sum of the coefficient of $x^{19}$ in $(1+x)^{49}$ and the coefficient of $x^{20}$ in $(1+x)^{49}$. Coefficient of $x^{19}$ in $(1+x)^{49}$ is ${}^{49}C_{19}$. Coefficient of $x^{20}$ in $(1+x)^{49}$ is ${}^{49}C_{20}$. Thus,the constant term is $31 \cdot {}^{49}C_{19} + {}^{49}C_{20}$. Using the identity ${}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}$,we have ${}^{49}C_{20} + {}^{49}C_{19} = {}^{50}C_{20}$. So,the constant term is $30 \cdot {}^{49}C_{19} + ({}^{49}C_{19} + {}^{49}C_{20}) = 30 \cdot {}^{49}C_{19} + {}^{50}C_{20}$.

The constant term in the expansion of $\left(1+\frac{1}{x}\right)^{20}\left(30 x(1+x)^{29}+(1+x)^{30}\right)$ is

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