Binomial Summations 6

Algebra Level 5

If the expansion of ( 1 + x + 2 x 2 ) 20 (1+x+2x^2)^{20} is equal to a 0 + a 1 x + a 2 x 2 + ⋯ + a 40 x 40 a_0 + a_1 x+ a_2 x^2 + \cdots + a_{40} x^{40} for constants a 0 , a 1 , … , a 40 a_0,a_1, \ldots , a_{40} , compute

⌈ 1 1 0 8 ∑ r = 0 19 a 2 r ⌉ . \left \lceil \dfrac1{10^8} \sum_{r=0}^{19} a_{2r} \right \rceil .


The answer is 5498.

Kishore S. Shenoy by Kishore S. Shenoy , 22, India

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2 solutions

( 1 + x + 2 x 2 ) 20 = a 0 + a 1 x + a 2 x 2 + ⋯ + a 40 x 40 Putting x = 1 4 20 = a 0 + a 1 + a 2 + ⋯ + a 40 Putting x = − 1 2 20 = a 0 − a 1 + a 2 − ⋯ + a 40 ⟹ 4 20 + 2 20 = 2 a 0 + 2 a 2 + 2 a 4 + ⋯ + 2 a 40 ⟹ ∑ r = 0 20 a 2 r = 4 20 + 2 20 2 ∑ r = 0 19 a 2 r = 2 39 + 2 19 − a 40 = 2 39 + 2 19 − 2 20 = 2 39 − 2 19 = 5.49755 × 1 0 11 ⟹ ⌈ 1 1 0 8 ∑ r = 0 19 a 2 r ⌉ = 5498 \begin{aligned} (1+x+2x^2)^{20} & = a_0 + a_1 x + a_2 x^2 + \cdots + a_{40} x^{40} & \small \color{#3D99F6}{\text{Putting }x=1} \\ 4^{20} & = a_0 + a_1 + a_2 + \cdots + a_{40} & \small \color{#3D99F6}{\text{Putting }x=-1} \\ 2^{20} & = a_0 - a_1 + a_2 - \cdots + a_{40} \\ \implies 4^{20} + 2^{20} & = 2a_0 + 2a_2 + 2a_4 + \cdots + 2a_{40} \\ \implies \sum_{r=0}^{20} a_{2r} & = \frac {4^{20} + 2^{20}}2 \\ \sum_{r=0}^{\color{#3D99F6}{19}} a_{2r} & = 2^{39}+2^{19} - a_{40} \\ & = 2^{39}+2^{19} - 2^{20} \\ & = 2^{39}-2^{19} \\ & = 5.49755 \times 10^{11} \\ \implies \left \lceil \frac 1{10^8} \sum_{r=0}^{19} a_{2r} \right \rceil & = \boxed{5498} \end{aligned}

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In my addition I made an error of 1 ! But my method was long.
N o t e t h a t a 2 r i s s u m o f s e v e r a l t e r m s . A l s o N ! a ! = N ! ( N − a ) ! ⌈ 1 1 0 8 ∑ r = 0 19 a 2 r ⌉ = ⌈ 1 1 0 8 ∑ r = 0 10 ∑ k = 0 r 20 ! ( r − k ) ! ∗ ( 2 ∗ k ) ! ∗ ( 20 − r − k ) ! ∗ ( 2 r − k + 2 20 − r − k ) ∗ ( 1 − ⌊ r 10 ⌋ ) ) ⌉ = 5498. ( 1 − ⌊ r 10 ⌋ ) i s u s e d s i n c e r = 10 i s m i d d l e t e r m h a v i n g n o p a i r i n g . Note~ that~a_{2r} ~is ~sum ~of~several ~terms.~~ Also~\dfrac{N!}{a!}=\dfrac{N!}{(N-a)!}\\ \displaystyle \left \lceil \dfrac1{10^8} \sum_{r=0}^{19} a_{2r} \right \rceil \\ \displaystyle = \left \lceil~~ \dfrac1 {10^8} \sum_{r=0}^{10} \displaystyle \sum_{k=0}^r \dfrac{20!}{(r-k)!*(2*k)!*(20-r-k)!}*\left (2^{r-k} +2^{20-r-k})*\left (1- \left \lfloor \dfrac r {10} \right \rfloor \right ) \right ) ~~ \right \rceil ~~=\color{#D61F06}{5498.} \\ \left (1- \left \lfloor \dfrac r {10} \right \rfloor \right ) ~~is ~used ~since~~ r=10~is~middle~term~having ~no ~pairing.

Niranjan Khanderia - 3 years, 2 months ago
Abu Zubair
Nov 4, 2015

PUT

X=1.................(1)

X=-1................(2)

THEN ADD (2) AND (1)

DIVIDE RESULT BY '2'

SUBTRACT a 40 a_{40} FROM IT(i.e 2 20 2^{20} )

divide it by 1 0 8 10^{8}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I Think It Is The Easiest Problem Of The Set And Also Has The Highest Rating !!

Prakhar Bindal - 5 years, 7 months ago

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Did the same way. Made a mistake in estimating the ceiling function.

Put my first two answers as 5496,5497 :/

Then put 5498 and got it right.

A Former Brilliant Member - 5 years, 6 months ago

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I put 5497, it showed incorrect.So i just gave it up!!!!1

Sathya NC - 5 years, 4 months ago

This is incorrect. You have only shown that it's true for x = ± 1 x=\pm 1 , when the question states that we're looking for all real x x .

Pi Han Goh - 5 years, 4 months ago

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