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  • If is the coefficient of <img alt="x^{10-r}" src="https://entrancecorner.oncodecogs.com/gif.latex?x%5E%7B10-r%7D"

If a_{r} is the coefficient of x^{10-r} in the Binomial expansion of (1+x)^{10},then \sum_{r=1}^{10} r^{3}\left(\frac{a_{r}}{a_{r-1}}\right)^{2}is equal to

Option: 1

5445


Option: 2

3025


Option: 3

4895


Option: 4

 1210


Answers (1)

best_answer

\sum_{\mathrm{r}=1}^{10} \mathrm{r}^{3}\left[\frac{\mathrm{a}_{\mathrm{r}}}{\mathrm{a}_{\mathrm{r}-1}}\right]^{2}
\because \frac{\mathrm{a}_{\mathrm{r}}}{\mathrm{a}_{\mathrm{r}-1}}=\frac{10-\mathrm{r}+1}{\mathrm{r}}=\frac{11-\mathrm{r}}{\mathrm{r}}            \because(1+\mathrm{x})^{10} \Rightarrow{ }^{10} \mathrm{C}_{\mathrm{r}} \mathrm{n}^{\mathrm{r}}
                                                                          \Rightarrow{ }^{10} \mathrm{C}_{10-\mathrm{r}} \mathrm{n}^{10-\mathrm{r}}
                                                                          ={ }^{10} \mathrm{C}_{10-\mathrm{r}} \text { or }={ }^{10} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{10-\mathrm{r}}
                                                                        \mathrm{a}_{\mathrm{r}}={ }^{10} \mathrm{C}_{\mathrm{r}}

\sum_{r=1}^{10} r^{3}\left[\frac{11-r}{r}\right]^{2}
\sum_{\mathrm{r}=1}^{10} r(11-r)^{2}
\sum_{\mathrm{r}=1}^{10}\left[r^{3}-22 r^{2}+121 r\right]
=\left(\frac{(10)(11)}{2}\right)^{2}-22\left(\frac{(10)(11)(21)}{6}\right)+\left(\frac{(10)(11)}{2}\right)(121)
=1210 \text { Ans. }

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