JEE Mains 2021 solution. Find middle term in expansion (x^2 + 1/x^2 +2)^n

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show that middle term in the expansion of [(1+x)^(2n) is 1.3.5... (2n-1)/n! ]2^n x^n where n is ... show that middle term in the expansion of [(1+x)^(2n) is 1.3.5... (2n-1)/n! ]2^n x^n where n is ...
If the middle term in the expansion of (x² + 1/x)" is 924 x⁶, then find the value of n. If the middle term in the expansion of (x² + 1/x)" is 924 x⁶, then find the value of n.
In expansion of (1+x)^2n middle term is ((1.3.5...(2n-1))*2^n*x^n)/n! | Binomial Theorem In expansion of (1+x)^2n middle term is ((1.3.5...(2n-1))*2^n*x^n)/n! | Binomial Theorem
The middle term in the expansion of (1+x)^2 n is (1) (2 n) !/n ! x^2 (2) (2 n) !/n !(n-1) ! x^n+1... The middle term in the expansion of (1+x)^2 n is (1) (2 n) !/n ! x^2 (2) (2 n) !/n !(n-1) ! x^n+1...
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Show that the middle term in the expansion `(x-1/x)^(2n)i s` `(1. 3. 5 (2n-1))/n(-2)^n` . Show that the middle term in the expansion `(x-1/x)^(2n)i s` `(1. 3. 5 (2n-1))/n(-2)^n` .
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Middle Term Of A Binomial Expansion | Prove: Middle Term Of (1+x)^2n is 1.3.5.....(2n-1).2^n.x^n/n! Middle Term Of A Binomial Expansion | Prove: Middle Term Of (1+x)^2n is 1.3.5.....(2n-1).2^n.x^n/n!