1
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Question

# 11C01 + 11C12 + 11C23+ .................... + 11C1011 =

A
211111
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B
21116
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C
411111
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D
311111
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Solution

## The correct option is B 211−16We can use binomial expansion to do this(1+x)11=11C0+11C1x+11C2x2.........11C10x10+11C11x11Now integrate the following code∫10(1+x)11=11C0+11C1x+11C2x2.........11C10x10+11C11x11→(1+x)1212−112=11C01+11C1x2+11C2x33.........11C10x1011+11C11x1212Now put x=1→(1+1)12−112=11C01+11C12+11C23.........11C1011+11C1112→(2)12−112−112=11C01+11C12+11C23.........11C1011Thus11C01+11C12+11C23.........11C1011=212−212=211−16

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