Binomial Theorem - 4

Illustration 8

If (3 + 4x)n = P0+ P1x + P2x2 +...+ Pnxn then prove that (Po - P2 + P4...)2 + (P1 - P3+ P5 ....)2 = 25n

Ans . Let us put x = i in the expansion

(3 + 4x)n = Po+ P1x + P2 x2 +.............+Pnxn

So, we get

(3 + 4i)n = (Po + P2 + P4 ...........) + i (P1 -P3 + P5 ...............)

Now equating the square of the modulus, we get

(Po + P2 + P4 ...........) 2+ (P1 -P3 + P5 ...............)2

= (32+ 42) n

= 25 n

Dumb Question Why did P4 got a +ve sign in front of it ?

Ans P4 is Coefficient of x4 . So,when we put i. i4 gives the sign and .

i4 = (i2)2 = (-1)2 = 1

Binomial Series (for negative or fractional indices)

If and | x | < 1, then

(1 + x)n = 1 + nx +

General term = T r+1

Illustration 9

Find the coefficients of x4andx-4 in expansion of

, | < x < 2 .



=

=



=

Coefficient of x4 = 0 +

Coefficient of x-4 =

Binomial Theoram

If n is a positive integer then

( a1 + a2 + ........+ am) n a1n1a2n2. a3n3.......am nm



Where the summation is taken over all non-negative integers n1, n2 ..........nm such that n1+ n2 + n3+.....+ nm = n

and the number of terms in the expansion is

= number of non-negative integral soln of equation

n1 + n2 +..........+nm

= n + m -1 Cm-1

Illustration 10 .

find the total number of terms in the kexpansion

of (x + y + z + w) n

Ans. From Hultinomial Theorm

(x + y + z +w )n = x1n1 y n2 zn3 wn4

where n1, n2, n3, n4 are non negative integers subject to the condition n1 + n2 + n3 + n4 = n

Hence no of distinct terms

= Coefficient of xnin (xo + x1 + x2 +.......+xn)4

= Coefficient of xn in

= Coefficient of x nin (1 - xn+1)4 (1 - x)-1

= Coefficient of x nin (1 - x)-4 (since xn+1

= n + 3Cn

=

Dumb Question Why is no of disfinft terms equal to coefficient

of xn in (xo + x1 + x2+....................+xn )4 ?

Ans This is actually a problem of permutation and combipnation

But Let us discuss it here very brifly .

Equation is n1 + n2 + n3 + n4 = n

Now n1 can vary from o to n

Similar is the case for n2, n3 and n4.

So, xo + x1 + x2 +..................+ xn for n1

and (xo + x1 + x2 +..................+ x)4 for all n1, n2 + n3 and n4.

And then we try to find coeft of xn as n1 n2+ n3 + n4 head to sum to n

Easy

(1)Prove that Co + C1 + C2+..........Cn = 2n

Ans we know

Co + C1 + C2+..........Cnxn = (1 + x) n

Now put x = 1 or both sides

Co + C1 + C2+..........Cn = 2n

(2) find the sum kof series 20 Cr ?

Ans We know = 20 Cr = 220

Now L.H.S. mhas 21lterms out of which 10 pairs are equal.

because nCr= nCn - r

So, L.H.S. 2 20 Cr - 20 C10 = 220

So, 2 = 2 20 + 20 C10

or (2 20 + 20 C10)


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