If - r -0 n n C r -1- n C r - n C r -13-25-24- then n is equal toA- 4B- 3C- 5D- 6

The correct option is C 5Let _r=^nC_r 1/^nC_r+^nC_r 1=^nC_r 1/^n+1C_r=^nC_r 1/n+1/r^nC_r 1 ∴ l_r=r/n+1 Now, S=∑^n_r=0{l_r}^3⇒ S=∑^n_r=0r^3/(n+1)^3=1/(n+1)^3∑^n ...

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Standard XI

Mathematics

Binomial Coefficients

If ∑ r =0 n n...

Question

If $n \sum r = 0 {\frac{^{n} C_{r - 1}}{^{n} C_{r} +^{n} C_{r - 1}}}^{3} = \frac{25}{24}$ ,then n is equal to

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Solution

The correct option is C 5
$L e t_{r} = \frac{^{n} C_{r - 1}}{^{n} C_{r} +^{n} C_{r - 1}}$ $= \frac{^{n} C_{r - 1}}{^{n + 1} C_{r}} = \frac{^{n} C_{r - 1}}{{\frac{n + 1}{r}}^{n} C_{r - 1}}$
$∴ l_{r} = \frac{r}{n + 1}$
$N o w,$
$S = n \sum r = 0 {l_{r}}^{3} \Rightarrow S = n \sum r = 0 \frac{r^{3}}{(n + 1)^{3}} = \frac{1}{(n + 1)^{3}} n \sum r = 0 r^{3}$
$\Rightarrow S = \frac{1}{(n + 1)^{3}} {\frac{n (n + 1)}{2}}^{2} \Rightarrow S = \frac{n^{2}}{4 (n + 1)}$
$N o w, S = \frac{25}{24} (g i v e n) w h i c h i s o n l y p o s s i b l e f o r 5$

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