Question

In the spread of (1 + x) ^ n, the coefficient of three consecutive positions is 36,84 and 126 and then the value of n is known ?

Answer 1

$\bold{\huge{\underline{Question}}}$

In the spread of (1 + x) ^ n, the coefficient of three consecutive positions is 36,84 and 126 and then the value of n is known ?





$\bold{\huge{\underline{solution-}}}$



Suppose$T _{r} \: , T_{r + 1}, T_{r + 2}$ are three consecutive terms, then



$^{n} C_{r - 1} \: = 36 \\ \\ \\ ^{n} C_{r } = 84 \\ \\ \\ ^{n} C_{r + 1} = 126$



But we know that $\bf ^{n} C_{r} = \frac{n - r + 1}{r} \times \: ^{n}C_{r - 1}$




$.°. \: \: \: \: \: \frac{n + 1}{r} - 1 = \frac{84}{7} = \frac{7}{3} \\ \\ \\ it \: \: means \: \: \frac{n + 1}{r} = \frac{10}{3} ..................(1)$





$And \: \: \: \: \: \: \: \frac{n + 1}{r + 1} - 1 = \frac{126}{84} = \frac{3}{2} \\ \\ it \: means \: \: \: \: \: \frac{n + 1}{r + 1} = \frac{5}{2} .......................(2)$





$\bold{Now,\:\: By\: splitting\: from \:( 2)\: \:into \:equation\: (1),}$



4r = 3 (r + 1) means, r = 3



so,



n = 9

Answer 2

HIII......

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