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TS EAMCET 9 May 2024 Shift 1 Solved Paper

Section: Mathematics

Question : 19 of 160

Marks: +1, -0

{}^n\mathrm{C}_r = \mathrm{C}_r

2\;\frac{\mathrm{C}_1}{\mathrm{C}_0} + 4\;\frac{\mathrm{C}_2}{\mathrm{C}_1} + 6\;\frac{\mathrm{C}_3}{\mathrm{C}_2} + \dots + 2n\;\frac{\mathrm{C}_n}{\mathrm{C}_{n-1}} = 650

{}^n\mathrm{C}_2 =

Solution:

We have,

2\;\frac{c_1}{c_0} + 4\;\frac{c_2}{c_1} + 6\;\frac{c_3}{c_2} + \dots + 2n\;\frac{c_n}{c_{n-1}} = 650

\;\Rightarrow\;\; \sum\limits_{r=1}^{n} 2r \frac{{}^n c_r}{n^n c_{r-1}} = 650

\;\Rightarrow\;\; 2 \sum\limits_{r=1}^{n} r \left[ \frac{ \frac{n!}{r!(n-r)!} }{ \frac{n!}{(r-1)!(n-r+1)!} } \right] = 650

\;\Rightarrow\; 2 \sum\limits_{r=1}^{n} r \left[ \begin{array}{c} r \cdot (r-1)! (n-r)! \times \; \frac{ (n!-1)!(n-r+1) \cdot (n-r)! }{ n! } \end{array} \right]

= 650

\;\Rightarrow\;\; 2 \Sigma (n-r+1) = 650

\;\Rightarrow\;\; 2 \left[ n^2 + n - \frac{n(n+1)}{2} \right] = 650

\;\Rightarrow\;\; n^2 + n - 650 = 0

\;\Rightarrow\;\; (n+26)(n-25) = 0

\;\Rightarrow\;\; n = 25

\;\text{ Hence, }\; {}^n c_2 = {}^25 c_2 = 300

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