Illustration 8

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

Ans . Let us put x = i in the expansion

(3 + 4x)ⁿ = P_o+ P₁x + P₂ x² +.............+P_nxⁿ

So, we get

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

Now equating the square of the modulus, we get

(P_o + 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 + ........+ a_m) ⁿ a₁ⁿ¹a₂ⁿ². a₃ⁿ³.......a_m ^nm



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

and the number of terms in the expansion is

= number of non-negative integral solⁿ of equation

n₁ + n₂ +..........+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 = 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 (x^o + 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 + 3_Cn

=

Dumb Question Why is no of disfinft terms equal to coefficient

of xⁿ in (x^o + 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, x^o + x¹ + x² +..................+ xⁿ for n₁

and (x^o + 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 C_o + C₁ + C₂+..........C_n = 2ⁿ

Ans we know

C_o + C₁ + C₂+..........C_nxⁿ = (1 + x) ⁿ

Now put x = 1 or both sides

C_o + C₁ + C₂+..........C_n = 2ⁿ

(2) find the sum kof series ^{20 C_r} ?

Ans We know = ²⁰ C_r = 2²⁰

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

because ⁿC_r= ⁿC_{n - r}

So, L.H.S. 2 ²⁰ C_r - ²⁰ C₁₀ = 2²⁰

So, 2 = 2 ²⁰ + ²⁰ C₁₀

or (2 ²⁰ + ²⁰ C₁₀)