If-n-is-positive-integer-prove-that-the-cofficient-of-x-2-and-x-3-in-the-expansion-of-x-2-2x-2-n-are-2-n-1-n-2-and-2-n-1-n-n-1-1-3-

Question Number 9573 by lepan last updated on 17/Dec/16

I f n i s p o s i t i v e i n t e g e r p r o v e t h a t

t h e c o f f i c i e n t o f x^{2} a n d x^{3} i n t h e

e x p a n s i o n o f {(x^{2} + 2 x + 2)}^{n} a r e 2^{n - 1} . n^{2}

a n d 2^{n - 1} n (n - 1) \frac{1}{3} .

Commented by sou1618 last updated on 17/Dec/16

a, b, c = 0, 1, 2, 3 \dots . m (, n) = 1, 2, 3 \dots .

{k x}^{m} = \sum_{(a, b, c)} {{(x^{2})}^{a} \times {(2 x)}^{b} \times 2^{c} \times_{n} C_{a} \times_{n - a} C_{b} \times_{n - a - b} C_{c}}

k = c o f f i c i e n t o f x^{m}

m = 2 a + b + 0 c

a + b + c = n

_{n - a - b} C_{c} =_{c} C_{c} = 1

/ / / / / /

[{k x}^{2}]

m = 2 \Rightarrow (a, b, c) =^{(1)} (1, 0, n - 1),^{(2)} (0, 2, n - 2)

(1) k_{1} x^{2} = {(x^{2})}^{1} \times {(2 x)}^{0} \times {(2)}^{n - 1} \times_{n} C_{1} \times_{n - 1} C_{0}

\Rightarrow k_{1} = 2^{n - 1} \times n

(2) k_{2} x^{2} = {(x^{2})}^{0} \times {(2 x)}^{2} \times 2^{n - 2} \times_{n} C_{0} \times_{n} C_{2}

\Rightarrow k_{2} = 2^{n} \times \frac{n (n - 1)}{2} = 2^{n - 1} (n^{2} - n)

\Rightarrow k = k_{1} + k_{2} = 2^{n - 1} (n^{2} - n + n)

= 2^{n - 1} n^{2}

/ / / / / /

[{k x}^{3}]

m = 3 \Rightarrow (a, b, c) =^{(1)} (1, 1, n - 2),^{(2)} (0, 3, n - 3)

(1) k_{1} x^{3} = {(x^{2})}^{1} \times {(2 x)}^{1} \times 2^{n - 2} \times_{n} C_{1} \times_{n - 1} C_{1}

\Rightarrow k_{1} = 2 \times 2^{n - 2} \times n \times (n - 1) = 2^{n - 1} (n^{2} - n)

(2) k_{2} x^{3} = {(x^{2})}^{0} \times {(2 x)}^{3} \times 2^{n - 3} \times_{n} C_{0} \times_{n} C_{3}

\Rightarrow k_{2} = 2^{3} \times 2^{n - 3} \times \frac{n (n - 1) (n - 2)}{6} = 2^{n - 1} \frac{(n^{2} - n) (n - 2)}{3}

\Rightarrow k = k_{1} + k_{2} = 2^{n - 1} (n^{2} - n) (1 + \frac{n - 2}{3})

= 2^{n - 1} \frac{(n - 1) n (n + 1)}{3}

Facebook Tweet Pin