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Algebra Level 5

Expanding ( 1 + 0.2 ) 1000 (1+0.2)^{1000} by the Binomial theorem gives

( 1000 0 ) ( 0.2 ) 0 + ( 1000 1 ) ( 0.2 ) 1 + ( 1000 2 ) ( 0.2 ) 2 + ⋯ + ( 1000 1000 ) ( 0.2 ) 1000 {1000 \choose 0}(0.2)^0+{1000 \choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000} = A 0 + A 1 + A 2 + ⋯ + A 1000 = A_0 + A_1 + A_2 + \cdots + A_{1000}

Where A k = ( 1000 k ) ( 0.2 ) k A_k = {1000 \choose k}(0.2)^k for k = 0 , 1 , 2 , … , 1000 k = 0,1,2,\ldots,1000 .

For which k k_{}^{} is A k A_k^{} the largest?

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The answer is 166.

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Parth Lohomi by Parth Lohomi , 20, India

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3 solutions

Pranjal Jain
Mar 17, 2015

A k + 1 A k > 1 ⇒ ( 1000 k + 1 ) × 0.2 ( 1000 k ) > 1 ⇒ 1000 − k k + 1 > 5 ⇒ k < 165.833 \frac{A_{k+1}}{A_{k}}>1\\\Rightarrow \frac{\dbinom{1000}{k+1}×0.2}{\dbinom{1000}{k}}>1\\\Rightarrow \frac{1000-k}{k+1}>5\\\Rightarrow k<165.833

Therefore, A 166 > A 165 A_{166}>A_{165} but A 167 ≯ A 166 A_{167}\ngtr A_{166} . Thus, maximum is A 166 A_{166}

10 Helpful 0 Interesting 0 Brilliant 0 Confused

Wouldn't say a level fiiiiive problem dude!! ...

jaiveer shekhawat - 6 years, 2 months ago

@Pranjal Jain , note that the bounds of your solution are wrong.

1000 − k k + 1 > 5 ⟹ 6 k < 995 ⟹ k < 165.833 ⟹ k + 1 < 166.833 \frac{1000-k}{k+1}\gt 5\implies 6k\lt 995\implies k\lt 165.833\implies k+1\lt 166.833

which implies (from the first statement of your solution) that A k + 1 A_{k+1} is strictly increasing till k + 1 = 166 k+1=166 or that A k A_k is strictly increasing till k = 166 k=166 and as such, we have, sup ⁡ ( { A k } ) = A 166 \sup(\{A_k\})=A_{166} .

Edit your solution accordingly and correct the bounds. Otherwise, it seems that the value of k k is unbounded above which is false.

Prasun Biswas - 6 years, 2 months ago

Is this not a highly over rated sum??Perfect for level 3 ..... even over rated for level 4 too...

rajdeep brahma - 3 years ago
Aniket Verma
Mar 18, 2015

I would like to share a generalised solution for these type of questions.

In the expansion of ( 1 + x ) n (1+x)^{n} , the greatest term is r.

Where r = [ p ] + 1 ( w h e r e [ . ] s t a n d s f o r G . I . F ) r=[p] + 1 ~ (where ~[.]~stands~for~G.I.F) , if p p is not an integer and r = p & p + 1 r= p~\& p+1 if p p is an integer.

N o t e , p = x ( 1 + N ) 1 + x Note,~p = \dfrac{x(1+N)}{1+x}

In given question x = 0.2 & n = 1000 x=0.2~ \&~ n=1000

therefore [ p ] = [ 0.2 ( 1001 ) 1.2 ] = 166 [p]=[\dfrac{0.2(1001)}{1.2}] = 166

hence 16 7 t h 167_{th} term is the greatest.

so, i n A k in~ A_{k} k = 166 k=166 . ( s i n c e s e r i e s s t a r t s f r o m A 0 ) (since ~series~ starts ~from~A_0)

7 Helpful 0 Interesting 0 Brilliant 0 Confused

from where did you learn that

Shashank Rustagi - 6 years, 2 months ago

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I have generalized it by observing the results of these types of questions by myself. later on i came to know that this result is really used in some books.

Aniket Verma - 6 years, 2 months ago

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great consequence of your observations Keep it up dude !!!

Shashank Rustagi - 6 years, 2 months ago

Good observation...Keep it up!!!!(+1)

rajdeep brahma - 3 years ago

0 Helpful 0 Interesting 0 Brilliant 0 Confused

These sums are meant to be done manually....and pls post solutions using mathematics...the topic is 'ALGEBRA' not 'COMPUTER SCIENCE'.

rajdeep brahma - 3 years ago

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