The-coefficient-of-x-5-in-the-expansion-of-1-x-21-1-x-22-1-x-30-is-

Question Number 56095 by gunawan last updated on 10/Mar/19

The coefficient of x^{5} in the expansion of

{(1 + x)}^{21} + {(1 + x)}^{22} + \dots + {(1 + x)}^{30} is

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Mar/19

{(1 + x)}^{n}

l e t (r + 1) t h t e r m c o n t a i n s x^{5}

{n c}_{r} x^{r} \to s o x^{r} = x^{5} \to r = 5

n o w {(1 + x)}^{21} \to 21 c_{5} x^{5}

{(1 + x)}^{22} \to 22 c_{5} x^{5}

\dots

\dots

r e q u i r e d a n s w e r i s

21 c_{5} + 22 c_{5} + 23 c_{5} + \dots + 30 c_{5}

Answered by mr W last updated on 10/Mar/19

{(1 + x)}^{21} + {(1 + x)}^{22} + \dots + {(1 + x)}^{30}

= \frac{{(1 + x)}^{21} [{(1 + x)}^{10} - 1]}{(1 + x) - 1}

= \frac{{(1 + x)}^{31} - {(1 + x)}^{21}}{x}

c o e f . o f x^{5} t e r m i s :

C_{6}^{31} - C_{6}^{21} = 682017

Commented by gunawan last updated on 10/Mar/19

wow thank you Sir

Commented by malwaan last updated on 10/Mar/19

w h a t i s t h e f o r m u l a

t h a t y o u u s e d ? p l e a s e

Commented by mr W last updated on 10/Mar/19

S = {(1 + x)}^{21} + {(1 + x)}^{22} + \dots + {(1 + x)}^{30}

i s a g e o m e t r i c p r o g r e s s i o n w i t h

a_{1} = {(1 + x)}^{21}

q = (1 + x)

n = 10

S = \frac{a_{1} (q^{n} - 1)}{q - 1} = \frac{{(1 + x)}^{21} [{(1 + x)}^{10} - 1]}{(1 + x) - 1}

= \frac{{(1 + x)}^{31} - {(1 + x)}^{21}}{x}

x^{6} t e r m o f {(1 + x)}^{31} i s C_{6}^{31} x^{6}

x^{6} t e r m o f {(1 + x)}^{21} i s C_{6}^{21} x^{6}

x^{5} t e r m o f S i s \frac{C_{6}^{31} x^{6} - C_{6}^{21} x^{6}}{x} = (C_{6}^{31} - C_{6}^{21}) x^{5}

Commented by malwaan last updated on 11/Mar/19

thank you sir

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