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  • The sum of the coefficients of three consecutive terms in the binomial expansion of <img alt="(1+\mathrm{x})^{\mathrm{n}+2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%281&plus;%5Cma

The sum of the coefficients of three consecutive terms in the binomial expansion of (1+\mathrm{x})^{\mathrm{n}+2}, which are in the ratio 1: 3: 5, is equal to

Option: 1

63


Option: 2

92


Option: 3

25


Option: 4

41


Answers (1)

best_answer

{ }^{\mathrm{n}+2} C_{r-1}:{ }^{\mathrm{n}+2} C_{r}:{ }^{\mathrm{n}+2} C_{r+1}:: 1: 3: 5

\frac{(n+2) !}{(r-1) !(n-r+3) !} \times \frac{r !(n+2-r) !}{(n+2) !}=\frac{1}{3}
\frac{\mathrm{r}}{(\mathrm{n}-\mathrm{r}+3)}=\frac{1}{3} \Rightarrow \mathrm{n}-\mathrm{r}+3=3 \mathrm{r}
n=4 r-3-0 \quad \dots\left ( 1 \right )

\frac{(n+1) !}{r !(n+2-r) !} \times \frac{(r+1) !(n-r+1) !}{(n+2) !}=\frac{3}{5}
\frac{r+1}{n+2-r}=\frac{3}{5}
8 r-1=3 n \quad \dots(2)

By equation 1 and 2

\frac{8 r-1}{3}=4 r-3 \quad \quad n=4 r-3
\mathrm{r}=2 \quad \quad \quad\, \mathrm{n}=4 \times 2-3
                            \mathrm{n}=5

Sum : { }^{7} \mathrm{C}_{1}+{ }^{7} \mathrm{C}_{2}+{ }^{7} \mathrm{C}_{3}=7+21+35=63

Posted by

Pankaj

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