Sum up these coefficients!

Probability Level 3

Find the sum of the following series :-

( C 0 − 2 2 × C 1 + 3 2 × C 2 − . . . . . . . + ( − 1 ) n ( n + 1 ) 2 × C n ) (^{ }{ C }_{ 0 }-{ 2 }^{ 2 }\times ^{ }{ C }_{ 1 }+{ 3 }^{ 2 }\times ^{ }{ C }_{ 2 }-.......+{ (-1 })^{ n }({ n+1 })^{ 2 }\times ^{ }{ C }_{ n }) where n>2.

D e t a i l s a n d A s s u m p t i o n s Details\quad and\quad Assumptions

C r = n C r ^{ }{ C }_{ r }=^{ n }{ C }_{ r }


The answer is 0.

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Aditya Tiwari by Aditya Tiwari , 22, India

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3 solutions

Yash Choudhary
Mar 1, 2015

⇒ ( 1 − x ) n = C 0 − C 1 x + C 2 x 2 − C 3 x 3 + . . . . ( − 1 ) n C n x n ⇒ x ( 1 − x ) n = C 0 x − C 1 x 2 + C 2 x 3 − C 3 x 4 + . . . . ( − 1 ) n C n x n + 1 D i f f e r e n t i a t e b o t h s i d e s w . r . t x , ⇒ ( 1 − x ) n + n x ( 1 − x ) n − 1 = C 0 − 2 C 1 x + 3 C 2 x 2 − 4 C 3 x 3 + . . . . ( − 1 ) n ( n + 1 ) C n x n ⇒ x ( ( 1 − x ) n + n x ( 1 − x ) n − 1 ) = C 0 x − 2 C 1 x 2 + 3 C 2 x 3 − 4 C 3 x 4 + . . . . ( − 1 ) n ( n + 1 ) C n x n + 1 A g a i n d i f f e r e n t i a t e b o t h s i d e s w . r . t x , ⇒ ( 1 − x ) n + n x ( 1 − x ) n − 1 + 2 n x ( 1 − x ) n − 1 + x 2 n ( n − 1 ) ( 1 − x ) n − 2 = C 0 − 2 2 C 1 x + 3 2 C 2 x 2 − 4 2 C 3 x 3 + . . . . ( − 1 ) n ( n + 1 ) 2 C n x n P u t t i n g x = 1 , w e g e t r e q u i r e d s e r i e s i n R . H . S a n d 0 i n L . H . S , H e n c e , a n s w e r i s 0 . \Rightarrow \quad { (1-x) }^{ n }={ C }_{ 0 }-C_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }-{ { C }_{ 3 }x }^{ 3 }+....{ (-1) }^{ n }{ C }_{ n }{ x }^{ n }\\ \Rightarrow \quad x{ (1-x) }^{ n }={ C }_{ 0 }x-C_{ 1 }{ x }^{ 2 }+{ C }_{ 2 }{ x }^{ 3 }-{ { C }_{ 3 }x }^{ 4 }+....{ (-1) }^{ n }{ C }_{ n }{ x }^{ n+1 }\\ Differentiate\quad both\quad sides\quad w.r.t\quad x,\\ \Rightarrow \quad { (1-x) }^{ n }+nx{ (1-x) }^{ n-1 }={ C }_{ 0 }-2C_{ 1 }x+3{ C }_{ 2 }{ x }^{ 2 }-4{ { C }_{ 3 }x }^{ 3 }+....{ (-1) }^{ n }(n+1){ C }_{ n }{ x }^{ n }\\ \Rightarrow \quad x({ (1-x) }^{ n }+nx{ (1-x) }^{ n-1 })={ C }_{ 0 }x-2C_{ 1 }{ x }^{ 2 }+3{ C }_{ 2 }{ x }^{ 3 }-4{ { C }_{ 3 }x }^{ 4 }+....{ (-1) }^{ n }(n+1){ C }_{ n }{ x }^{ n+1 }\\ Again\quad differentiate\quad both\quad sides\quad w.r.t\quad x,\\ \Rightarrow \quad { (1-x) }^{ n }+nx{ (1-x) }^{ n-1 }+2nx{ (1-x) }^{ n-1 }+{ x }^{ 2 }n(n-1){ (1-x) }^{ n-2 }={ C }_{ 0 }-{ 2 }^{ 2 }C_{ 1 }x+{ 3 }^{ 2 }{ C }_{ 2 }{ x }^{ 2 }-{ 4 }^{ 2 }{ { C }_{ 3 }x }^{ 3 }+....{ (-1) }^{ n }{ (n+1) }^{ 2 }{ C }_{ n }{ x }^{ n }\\ Putting\quad x=1,\quad we\quad get\quad required\quad series\quad in\quad R.H.S\quad and\quad 0\quad in\quad L.H.S,\\ Hence,\quad answer\quad is\quad \boxed { 0 } .

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Rohan Bansal
Mar 1, 2015

Well u can just evaluate by putting any value of n>2

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Aaghaz Mahajan
Apr 24, 2018

Well......simple JEE approach by keeping n=3

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