How to find the greatest absolute term in a binomial expansion
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I know that the ratio of consecutive terms should be ≥1 and I'm able to solve using that approach but I was wondering whether we could derive a general formula and how would it work?
binomial-theorem
asked Dec 14 '18 at 15:25
Sameer ThakurSameer Thakur
305
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add a comment |
$begingroup$
I know that the ratio of consecutive terms should be ≥1 and I'm able to solve using that approach but I was wondering whether we could derive a general formula and how would it work?
binomial-theorem
asked Dec 14 '18 at 15:25
Sameer ThakurSameer Thakur
305
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$begingroup$
Have you read this math.stackexchange.com/questions/493903/… ?
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– mm-crj
Dec 14 '18 at 15:30
$begingroup$
Thanks but I want to find the greatest absolute term not the greatest coefficient
$endgroup$
– Sameer Thakur
Dec 14 '18 at 16:18
add a comment |
$begingroup$
I know that the ratio of consecutive terms should be ≥1 and I'm able to solve using that approach but I was wondering whether we could derive a general formula and how would it work?
binomial-theorem
asked Dec 14 '18 at 15:25
Sameer ThakurSameer Thakur
305
$endgroup$
I know that the ratio of consecutive terms should be ≥1 and I'm able to solve using that approach but I was wondering whether we could derive a general formula and how would it work?
binomial-theorem
binomial-theorem
asked Dec 14 '18 at 15:25
Sameer ThakurSameer Thakur
305
asked Dec 14 '18 at 15:25
Sameer ThakurSameer Thakur
305
asked Dec 14 '18 at 15:25
Sameer ThakurSameer Thakur
305
asked Dec 14 '18 at 15:25
Sameer ThakurSameer Thakur
305
asked Dec 14 '18 at 15:25
Sameer ThakurSameer Thakur
305
305
$begingroup$
Have you read this math.stackexchange.com/questions/493903/… ?
$endgroup$
– mm-crj
Dec 14 '18 at 15:30
$begingroup$
Thanks but I want to find the greatest absolute term not the greatest coefficient
$endgroup$
– Sameer Thakur
Dec 14 '18 at 16:18
add a comment |
$begingroup$
Have you read this math.stackexchange.com/questions/493903/… ?
$endgroup$
– mm-crj
Dec 14 '18 at 15:30
$begingroup$
Thanks but I want to find the greatest absolute term not the greatest coefficient
$endgroup$
– Sameer Thakur
Dec 14 '18 at 16:18
$begingroup$
Have you read this math.stackexchange.com/questions/493903/… ?
$endgroup$
– mm-crj
Dec 14 '18 at 15:30
$begingroup$
Have you read this math.stackexchange.com/questions/493903/… ?
$endgroup$
– mm-crj
Dec 14 '18 at 15:30
$begingroup$
Thanks but I want to find the greatest absolute term not the greatest coefficient
$endgroup$
– Sameer Thakur
Dec 14 '18 at 16:18
$begingroup$
Thanks but I want to find the greatest absolute term not the greatest coefficient
$endgroup$
– Sameer Thakur
Dec 14 '18 at 16:18
add a comment |
1 Answer
1
active
oldest
votes
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My instantaneous thought on this is that the first binomial coefficient proceeds from the zeroth (1) by multiplying it by $n$; and the second by multiplying that by $(n-1)/2$, ... , & thereafter by $(n-k)/(k+1)$; and you're also multiplying by $x$ each time ... so the maximum term will be at the stationary point, when $(n-k)x=k+1$, and the increase 'turns round' to become a decrease. So you would have $$k≈frac{nx-1}{x+1} ,$$ the "≈" taking account of the fact that it goes in integer steps, & $x$ need not be an integer.
answered Dec 14 '18 at 17:55
AmbretteOrriseyAmbretteOrrisey
54210
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
My instantaneous thought on this is that the first binomial coefficient proceeds from the zeroth (1) by multiplying it by $n$; and the second by multiplying that by $(n-1)/2$, ... , & thereafter by $(n-k)/(k+1)$; and you're also multiplying by $x$ each time ... so the maximum term will be at the stationary point, when $(n-k)x=k+1$, and the increase 'turns round' to become a decrease. So you would have $$k≈frac{nx-1}{x+1} ,$$ the "≈" taking account of the fact that it goes in integer steps, & $x$ need not be an integer.
answered Dec 14 '18 at 17:55
AmbretteOrriseyAmbretteOrrisey
54210
$endgroup$
add a comment |
$begingroup$
My instantaneous thought on this is that the first binomial coefficient proceeds from the zeroth (1) by multiplying it by $n$; and the second by multiplying that by $(n-1)/2$, ... , & thereafter by $(n-k)/(k+1)$; and you're also multiplying by $x$ each time ... so the maximum term will be at the stationary point, when $(n-k)x=k+1$, and the increase 'turns round' to become a decrease. So you would have $$k≈frac{nx-1}{x+1} ,$$ the "≈" taking account of the fact that it goes in integer steps, & $x$ need not be an integer.
answered Dec 14 '18 at 17:55
AmbretteOrriseyAmbretteOrrisey
54210
$endgroup$
add a comment |
$begingroup$
My instantaneous thought on this is that the first binomial coefficient proceeds from the zeroth (1) by multiplying it by $n$; and the second by multiplying that by $(n-1)/2$, ... , & thereafter by $(n-k)/(k+1)$; and you're also multiplying by $x$ each time ... so the maximum term will be at the stationary point, when $(n-k)x=k+1$, and the increase 'turns round' to become a decrease. So you would have $$k≈frac{nx-1}{x+1} ,$$ the "≈" taking account of the fact that it goes in integer steps, & $x$ need not be an integer.
answered Dec 14 '18 at 17:55
AmbretteOrriseyAmbretteOrrisey
54210
$endgroup$
My instantaneous thought on this is that the first binomial coefficient proceeds from the zeroth (1) by multiplying it by $n$; and the second by multiplying that by $(n-1)/2$, ... , & thereafter by $(n-k)/(k+1)$; and you're also multiplying by $x$ each time ... so the maximum term will be at the stationary point, when $(n-k)x=k+1$, and the increase 'turns round' to become a decrease. So you would have $$k≈frac{nx-1}{x+1} ,$$ the "≈" taking account of the fact that it goes in integer steps, & $x$ need not be an integer.
answered Dec 14 '18 at 17:55
AmbretteOrriseyAmbretteOrrisey
54210
answered Dec 14 '18 at 17:55
AmbretteOrriseyAmbretteOrrisey
54210
answered Dec 14 '18 at 17:55
AmbretteOrriseyAmbretteOrrisey
54210
answered Dec 14 '18 at 17:55
AmbretteOrriseyAmbretteOrrisey
54210
54210
add a comment |
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$begingroup$
Have you read this math.stackexchange.com/questions/493903/… ?
$endgroup$
– mm-crj
Dec 14 '18 at 15:30
$begingroup$
Thanks but I want to find the greatest absolute term not the greatest coefficient
$endgroup$
– Sameer Thakur
Dec 14 '18 at 16:18