In the expansion of ( 3 x + 2 ) 2 0 1 6 , what is the exponent of x in the term with the largest coefficient?
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Wow,thank you!
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@rohit udaiwal I hope that was well explained.
General problem : For what power x k does the expansion of ( a x + b ) n have the largest coefficient?
Solution : The coefficient of x k is C k = ( n k ) a k b n − k = k ⋯ 1 n ⋯ ( n − k + 1 ) a k b n − k . This coefficient is greater than the previous one iff C k > C k − 1 k ⋯ 1 n ⋯ ( n − k + 1 ) a k b n − k > ( k − 1 ) ⋯ 1 n ⋯ ( n − k + 2 ) a k − 1 b n − k + 1 k n − k + 1 > a b k n + 1 > a b + a k < ( n + 1 ) b + a a . The greatest coefficient corresponds to the greatest integer k for which this is the case, so k = ⌊ ( n + 1 ) b + a a ⌋ .
In this case: k = ⌊ 2 0 1 7 ⋅ 5 3 ⌋ = 1 2 1 0 .
Interesting fact: If we divide the entire expression by 5 2 0 1 6 we get ( 5 3 x + 5 2 ) 2 0 1 6 , in which the coefficient of x k corresponds to the probability of having k successes in an 2 0 1 6 -times repeated experiment with success probability p = 5 3 . The "peak" of this distribution, which corresponds to the greatest exponent, is the expectation value of this distribution, E = n p = 2 0 1 6 ⋅ 5 3 ≈ 1 2 1 0 . This is how I originally guessed the answer to this question. Of course, the derivation above is more "solid" than this intuitive approach.
Nice way to bring in the Bernoulli trials.
I have a doubt, shouldn't it be,
k
≤
b
+
a
(
n
+
1
)
a
??
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If the equal sign applies, then coefficients C k and C k − 1 are equal.
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Yes. I only pointed that out, because in the next step you mentioned the floor function, and equality would hold if ( b + a ) ∣ a ( n + 1 )
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@A Former Brilliant Member – I had realized that but didn't want to clog up by solution. :)
There are nice solutions provided by both @Vighnesh Shenoy and @Manuel Kahayon. I only wanted to show the generalization to this problem by treating everything symbolically :
What is the exponent of x that have the greatest coefficient in the expansion of ( a x + b ) N ?
The approach is not far from how you would solve the previous problem, you will arrive at
⌊ a + b a ( N + 1 ) ⌋
What we need to be aware of is the sign of a or b . If exactly one of them is negative, we have to check if their corresponding index is even. If not, we have to check which neighboring coefficient is larger. In the case where we introduced negatives, it is necessary to change the expression to :
⌊ ∣ a ∣ + ∣ b ∣ ∣ a ∣ ( N + 1 ) ⌋
Did you mean, "What is the exponent of x in greatest term of the expansion ( a x + b ) N ?" Nice generalisation.
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Oops, thanks for the notice.
Similar to what @Akhil Bansal wrote with a tweak to it.
Since the first term is ( 3 x ) 2 0 1 6 , and the nth term is equal to ( 3 ) ( n − 1 ) ( n − 1 t h t e r m ) ( 2 0 1 8 − n ) ( 2 ) , then as long as ( 3 ) ( n − 1 ) ( 2 0 1 8 − n ) ( 2 ) ≥ 1 , then the nth term will be greater than the n-1th term. Equating this, we get
( 3 ) ( n − 1 ) ( 2 0 1 8 − n ) ( 2 ) ≥ 1
8 0 7 . 8 ≥ n
Rounding off, the 807th term has the largest coefficient, meaning, the 807th term has 1 2 1 0 is the exponent of x in the 807th term.
We can directly use the formula,
Numerically greatest term in the expansion of
(
1
+
x
)
n
is
1
+
∣
x
∣
(
n
+
1
)
∣
x
∣
=
m
where
⌊
m
⌋
is exponent of x
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Oh, thanks, didn't know that...
Sorta disappointed to know that there was already a formula for this...
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@Akhil Bansal The formula you provided gives the numerically greatest term, that is, along with the input for value of x. But, the question asked the term with greatest co-efficient which may or may not be the same.
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@A Former Brilliant Member – Yes, formula also tells about which term contains numerically greatest term.That's how I finded exponent of x but I made one assumption that x = 1.
Why is it not 2016/2=1008,and can you provide a proof of this formula?
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Well, I do know that if it is the expansion of ( n x + n ) 2 0 1 6 for some n then the 1008th term would be the greatest ( x C 2 x is the greatest for some x), but since there are different coefficients (i.e. ( a x + b ) 2 0 1 6 , the terms with the largest coefficient tends to "shift" towards the left or the right, either left if a is greater than b or vice versa.
(Sorry, i kinda suck at explaining)
(Also, why is this in combinatorics not algebra?)
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@Manuel Kahayon – It is in combinatorics because it involved the binomial co-efficients.
it doesn't satisfy the inequalities: 2n+2>5k and 2n-3<5k . 807 is the only value that do
Sorry, I didn't know the proof. I studies it in 2014.
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Let T r denote the co-efficient of r t h term in the expansion of ( 3 x + 2 ) 2 0 1 6
T r + 1 = ( r 2 0 1 6 ) 3 2 0 1 6 − r 2 r
Replacing r by r-1,
T r = ( r − 1 2 0 1 7 − r ) 3 2 0 1 7 − r 2 r − 1
T r T r + 1 = 3 r 2 ( 2 0 1 7 − r ) . . . ( 1 )
Solving for T r T r + 1 ≥ 1 yields r ≤ 5 4 0 3 4
∴ r ≤ 8 0 6 . 8
Replacing r by r+1 in ( 1 ) ,
T r + 1 T r + 2 = 3 ( r + 1 ) 2 ( 2 0 1 6 − r )
Solving for T r + 1 T r + 2 ≤ 1 yields r ≥ 5 4 0 2 9
∴ r ≥ 8 0 5 . 8
The only integer satisfying both conditions is 806.
∴ r = 8 0 6
Hence the greatest co-efficient term is ( r + 1 ) = 8 0 6 + 1 = 8 0 7 term.
Co-efficient of x in 8 0 7 t h term = 2 0 1 6 − 8 0 6 = 1 2 1 0