Binomial Approximation

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binomial approximation, binomial, binomial expansion, approximation, binomial theorem, normal approximation to binomial probability, binomial distribution, solving binomial approximation, normal approximation, approximation of binomial theorem, binomial expansion of approximation, binomial expansion for approximation, expansion of binomial approximations, binomial expansion approximations, binomial distribution normal approximation, normal approximation to binomial distribution
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Set a:= 623^( 1/4 ). Then we have
2 = 625 - 623
= 5^4 - a^4
= ( 5^2 + a^2 )( 5 + a )( 5 - a )
By using 4 < a < 5, we have
( 5^2 + 4^2 )( 5 + 4 )( 5 - a ) < 2 < ( 5^2 + 5^2 )( 5 + 5 )( 5 - a )
Then we get approximated value of a as
5 - 2/369 < a < 5 - 1/250

According to sharper estimation, we may use
( 5^2 + 4^2 )( 5 + a )( 5 - a ) < 2 < ( 5^2 + 5^2 )( 5 + a )( 5 - a )

田村博志-zy
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You really cleared my doubt
You are 🙏🙏

kumkum
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Bro I feel very bad for u
U are very under rated
Bro u deserve a million subs

praveenjain
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starting cleared but last one is not clear

theamanshortss
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1000^1/3 is equal to 10
How????
And
625^1/4 is equal to 5

SHRISTYKUMARI-wyrw
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Rataa marke aaye ho kya bhai... bas direct amswer de rhe ho

srijalupadhyay
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I don't understand how you cut and how you write 999/1000 in 3:55 😭

djsniper
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How 999/1000 is coming I didn't understand

vishnukesarwani