The coefficient of three consecutive terms in the expansion of $(1 + x)^{n}$ are in the ratio 1 : 7 : 42 them prove that n = 55

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Solution

Let the successive terms be $T_{r - 1}, T_{r}, T_{r + 1}$ whose coefficient are 1 , 7 , and 42 respectively

$\frac{C_{r}}{C_{r - 1}} = \frac{n - r + 1}{r}$
$∴ \frac{n - r + 1}{r} = \frac{7}{1}$
$∴ n - 8 r + 1 = 0$

$\frac{C_{r + 1}}{C_{r}} = \frac{n - r}{r + 1}$
$∴ \frac{n - r}{r + 1} = \frac{42}{7} = \frac{6}{1}$
$∴ n - 7 r - 6 = 0$

$r = 7$
$∴ n = 55$

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The coefficient of three consecutive terms in the expansion of (1+x)n are in the ratio 1 : 7 : 42 them prove that n = 55

Let the successive terms be Tr−1,Tr,Tr+1 whose coefficient are 1 , 7 , and 42 respectivelyCrCr−1=n−r+1r ∴n−r+1r=71∴n−8r+1=0Cr+1Cr=n−rr+1 ∴n−rr+1=427=61∴n−7r−6=0r=7∴n=55

The coefficient of three consecutive terms in the expansion of $(1 + x)^{n}$ are in the ratio 1 : 7 : 42 them prove that n = 55