# Binomial Theorem - 4

__Illustration 8__

If (3 + 4x)

^{n}= P

_{0}+ P

_{1}x + P

_{2}x

^{2}+...+ P

_{n}x

^{n}then prove that (P

_{o}- P

_{2}+ P

_{4}...)

^{2 }+ (P

_{1}- P

_{3}+ P

_{5}....)

^{2}= 25

^{n}

__Ans__. Let us put x = i in the expansion

(3 + 4x)

^{n}= P

_{o}+ P

_{1}x + P

_{2}x

^{2}+.............+P

_{n}x

^{n}

So, we get

(3 + 4i)

^{n}= (P

_{o}+ P

_{2}+ P

_{4}...........) + i (P

_{1}-P

_{3}+ P

_{5}...............)

Now equating the square of the modulus, we get

(P

_{o}+ P

_{2}+ P

_{4}...........)

^{2}+ (P

_{1}-P

_{3}+ P

_{5}...............)

^{2}

= (3

^{2}+ 4

^{2})

^{n}

= 25

^{n}

__Dumb Question__Why did P

_{4}got a +ve sign in front of it ?

__Ans__P

_{4}is Coefficient of x

^{4}. So,when we put i. i

^{4}gives the sign and .

i

^{4}= (i

^{2})

^{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 x

^{4}andx

^{-4}in expansion of

, | < x < 2 .

=

=

=

Coefficient of x

^{4}= 0 +

Coefficient of x

^{-4}=

__Binomial Theoram__

If n is a positive integer then

( a

_{1}+ a

_{2}+ ........+ a

_{m})

^{n}a

_{1}

^{n1}a

_{2}

^{n2}. a

_{3}

^{n3}.......a

_{m}

^{nm}

Where the summation is taken over all non-negative integers n

_{1}, n

_{2}..........n

_{m}such that n

_{1}+ n

_{2}+ n

_{3}+.....+ n

_{m}= n

and the number of terms in the expansion is

= number of non-negative integral sol

^{n}of equation

n

_{1}+ n

_{2}+..........+n

_{m}

=

^{n + m -1 }C

_{m-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}= x

_{1}

^{n1}y

^{n2}z

^{n3}w

^{n4}

where n

_{1}, n

_{2}, n

_{3}, n

_{4}are non negative integers subject to the condition n

_{1}+ n

_{2}+ n

_{3}+ n

_{4}= n

Hence no of distinct terms

= Coefficient of x

^{n}in (x

^{o}+ x

^{1}+ x

^{2}+.......+x

^{n})

^{4}

= Coefficient of x

^{n}in

= Coefficient of x

^{n}in (1 - x

^{n+1})

^{4}(1 - x)

^{-1}

= Coefficient of x

^{n}in (1 - x)

^{-4}(since x

^{n+1}

= n + 3

_{Cn}

=

__Dumb Question__Why is no of disfinft terms equal to coefficient

of x

^{n}in (x

^{o}+ x

^{1}+ x

^{2}+....................+x

^{n})

^{4}?

__Ans__This is actually a problem of permutation and combipnation

But Let us discuss it here very brifly .

Equation is n

_{1}+ n

_{2}+ n

_{3}+ n

_{4}= n

Now n

_{1}can vary from o to n

Similar is the case for n

_{2}, n

_{3}and n

_{4}.

So, x

^{o}+ x

^{1}+ x

^{2}+..................+ x

^{n}for n

_{1}

and (x

^{o}+ x

^{1}+ x

^{2}+..................+ x)

^{4}for all n

_{1}, n

_{2}+ n

_{3}and n

_{4}.

And then we try to find coeft of x

^{n}as n

_{1}n

_{2}+ n

_{3}+ n

_{4}head to sum to n

__Easy__

(1)Prove that C

_{o}+ C

_{1}+ C

_{2}+..........C

_{n}= 2

^{n}

__Ans__we know

C

_{o}+ C

_{1}+ C

_{2}+..........C

_{n}x

^{n}= (1 + x)

^{n}

Now put x = 1 or both sides

C

_{o}+ C

_{1}+ C

_{2}+..........C

_{n}= 2

^{n}

(2) find the sum kof series

^{20 Cr}?

__Ans__We know =

^{20}C

_{r}= 2

^{20}

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

because

^{n}C

_{r}=

^{n}C

_{n - r}

So, L.H.S. 2

^{20}C

_{r}-

^{20}C

_{10}= 2

^{20}

So, 2 = 2

^{20}+

^{20}C

_{10}

or (2

^{20}+

^{20}C

_{10})