The coefficient of three consecutive terms in the expansion of $(1 + x)^{k}$ are in the ratio $1 : 7 : 42$ . Then $k$ is equal to

55

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50

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54

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42

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Solution

The correct option is A $55$
Let $(r + 1), (r + 2)$ and $(r + 3)$ be 3 consecutive
terms then
$^{n} C_{r} :^{n} C_{r + 1}^{n} C_{r + 1} = 1 : 7 : 42$
$\frac{^{n} C_{r + 1}}{^{n} C_{r + 1}} = \frac{1}{7} \Rightarrow \frac{r + 1}{n - r} = 1 / 7 \Rightarrow n - 8 r = 7$
$\frac{^{n} C_{r + 1}}{^{n} C_{r + 2}} = 7 / 42 \Rightarrow \frac{r + 2}{n - r - 1} = 1 / 6 \Rightarrow n - 7 r = 13$
On solving
$n = 55$

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The coefficient of three consecutive terms in the expansion of (1+x)k are in the ratio 1:7:42. Then k is equal to

The correct option is A 55Let (r+1),(r+2) and (r+3) be 3 consecutiveterms thennCr:nCr+1nCr+1=1:7:42nCr+1nCr+1=17⇒r+1n−r=1/7⇒n−8r=7nCr+1nCr+2=7/42⇒r+2n−r−1=1/6⇒n−7r=13On solvingn=55

The coefficient of three consecutive terms in the expansion of $(1 + x)^{k}$ are in the ratio $1 : 7 : 42$ . Then $k$ is equal to