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Question

If the 21st and 22nd terms in the expansion of (1x)44 are equal, then the value of |8x| is 


Solution

Here, T22=T21
 44C21(x)21= 44C20(x)20
44C2144C20=1x
44!21!×23!×20!×24!44!=1x
2421=1x
8x=7|8x|=7

Mathematics

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