Question:
If \(\sum_{r=1}^{9}\left(\frac{r+3}{2^{r}}\right){}^{9}\mathrm{C}_{r}=\alpha\left(\frac{3}{2}\right)^{9}+\beta\), then \(\alpha+\beta\) is equal to :
If \(\sum_{r=1}^{9}\left(\frac{r+3}{2^{r}}\right){}^{9}\mathrm{C}_{r}=\alpha\left(\frac{3}{2}\right)^{9}+\beta\), then \(\alpha+\beta\) is equal to :
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Split the sum into \(\sum r\,{}^9C_r/2^r\) and \(3\sum {}^9C_r/2^r\) and use the binomial expansion of \((1+x)^9\) at \(x=\frac12\).
Updated On: Nov 12, 2026
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The Correct Option is C
Solution and Explanation
Write the sum as \(\sum_{r=1}^{9} r\,\frac{{}^9C_r}{2^r}+3\sum_{r=1}^{9}\frac{{}^9C_r}{2^r}\).
From \((1+x)^9=\sum {}^9C_r x^r\), differentiating gives \(\sum r\,{}^9C_r x^{r-1}=9(1+x)^8\), so \(\sum r\,{}^9C_r x^r=9x(1+x)^8\).
At \(x=\frac12\): \(\sum r\,{}^9C_r/2^r=\frac92\left(\frac32\right)^8=3\left(\frac32\right)^9\).
Also \(\sum_{r=1}^{9}{}^9C_r/2^r=\left(\frac32\right)^9-1\), so the second part is \(3\left(\frac32\right)^9-3\).
Total \(=6\left(\frac32\right)^9-3\), so \(\alpha=6,\ \beta=-3\) and \(\alpha+\beta=3\).
Answer: (3) 3
From \((1+x)^9=\sum {}^9C_r x^r\), differentiating gives \(\sum r\,{}^9C_r x^{r-1}=9(1+x)^8\), so \(\sum r\,{}^9C_r x^r=9x(1+x)^8\).
At \(x=\frac12\): \(\sum r\,{}^9C_r/2^r=\frac92\left(\frac32\right)^8=3\left(\frac32\right)^9\).
Also \(\sum_{r=1}^{9}{}^9C_r/2^r=\left(\frac32\right)^9-1\), so the second part is \(3\left(\frac32\right)^9-3\).
Total \(=6\left(\frac32\right)^9-3\), so \(\alpha=6,\ \beta=-3\) and \(\alpha+\beta=3\).
Answer: (3) 3
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