If -i-120 20Ci-120Ci- 20Ci-13-k21- then k equals -

(^20C_i 1^20C_i+ ^20C_i 1)=(^20C_i 1^21C_i) ∵ ^nC_r+ ^nC_r 1= ^n+1C_r ⇒(20!(i 1)!·(20 i+1))·(i!(21 i)!21!)=i21 ∴∑_i=1^20(^20C_i 1^20C_i+^20C_i 1)^3=∑_i=1^20 ...

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Byju's Answer

Standard XII

Mathematics

nCr Definitions and Properties

If ∑i=120 20C...

Question

If $20 \sum i = 1 {(\frac{^{20} C_{i - 1}}{^{20} C_{i} +^{20} C_{i - 1}})}^{3} = \frac{k}{21},$ then $k$ equals :

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Solution

The correct option is B $100$
$(\frac{^{20} C_{i - 1}}{^{20} C_{i} +^{20} C_{i - 1}}) = (\frac{^{20} C_{i - 1}}{^{21} C_{i}})$

$∵^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$

$\Rightarrow (\frac{20!}{(i - 1)! \cdot (20 - i + 1)}) \cdot (\frac{i! (21 - i)!}{21!}) = \frac{i}{21}$

$∴ 20 \sum i = 1 {(\frac{^{20} C_{i - 1}}{^{20} C_{i} +^{20} C_{i - 1}})}^{3} = 20 \sum i = 1 {(\frac{i}{21})}^{3}$
$= \frac{1}{(21)^{3}} {(\frac{20 \times (20 + 1)}{2})}^{2}$
$= \frac{100}{21}$
Hence, $k = 100$

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If 20∑i=1( 20Ci−120Ci+ 20Ci−1)3=k21, then k equals :

The correct option is B 100(20Ci−120Ci+ 20Ci−1)=(20Ci−121Ci) ∵ nCr+ nCr−1= n+1Cr ⇒(20!(i−1)!⋅(20−i+1))⋅(i!(21−i)!21!)=i21 ∴20∑i=1(20Ci−120Ci+20Ci−1)3=20∑i=1(i21)3 =1(21)3(20×(20+1)2)2 =10021 Hence, k=100

If $20 \sum i = 1 {(\frac{^{20} C_{i - 1}}{^{20} C_{i} +^{20} C_{i - 1}})}^{3} = \frac{k}{21},$ then $k$ equals :