First slide

Binomial theorem for positive integral Index

Question

For natural numbers m, n, if $(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + \dots$ $and a_{1} = a_{2} = 10, then (m, n) is$

Moderate

A

35,20
B

45,35
C

35,45
D

20,65

Solution

$(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + a_{3} y^{3} + \dots$ On differentiating w.r.t. y, we get $- m (1 - y)^{m - 1} (1 + y)^{n} + (1 - y)^{m} n (1 + y)^{n - 1}$
$= a_{1} + 2 a_{2} y + 3 a_{3} y^{2} + \dots - - - - (i)$
On putting y =0 in Eq. (i), we get $- m + n = a_{1} = 10$ $[∵ a_{1} = 10, given] \dots ..(ii)$
On again differentiating Eq. (i), we get
$\begin{aligned} - m [- (m - 1) (1 - y)^{m - 2} (1 + y)^{n} + (1 - y)^{m - 1} n (1 + y)^{n - 1}] \\ + n [- m (1 - y)^{m - 1} (1 + y)^{n - 1} + (1 - y)^{m} (n - 1) (1 + y)^{n - 2}] \end{aligned}$
$= 2 a_{2} + 6 a_{3} y + \dots . . . . . . . . . (iii)$
On putting Y= 0 in Eq. (iii), we get
$\begin{array}{l} - m [- (m - 1) + n] + n [- m + (n - 1)] = 2 a_{2} = 20 \\ \Rightarrow m (m - 1) - mn - mn + n (n - 1) = 20 \\ \Rightarrow m^{2} + n^{2} - m - n - 2 mn = 20 \\ \Rightarrow (m - n)^{2} - (m + n) = 20 \\ \Rightarrow 100 - (m + n) = 20 \\ \Rightarrow m + n = 80 - - - - - - - (iv) \end{array}$
On solving Eqs. (ii) and (iv), we get
m=35and n=45

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