k = 0 ∑ 1 0 ( − 1 ) k ( 2 k 2 0 ) = ?
Notation: ( N M ) denotes the binomial coefficient , ( N M ) = N ! ( M − N ) ! M ! .
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my solution is same too sir !! can we do it by another method ?
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I don't know a better method. I thought you used a different one. I also wanted to show a proper LaTex solution. I edited the problem for you.
plz edit the problem correctly @Chew-Seong Cheong
hello shubham dhull
man missing you on slack
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S = k = 0 ∑ 2 0 ( − 1 ) k ( 2 k 2 0 ) Since ( 2 k 2 0 ) = 0 for 2 k > 2 0 = k = 0 ∑ 1 0 ( − 1 ) k ( 2 k 2 0 ) = ( 0 2 0 ) − ( 2 2 0 ) + ( 4 2 0 ) − ( 6 2 0 ) + . . . + ( 2 0 2 0 )
We note that:
( 1 + i ) 2 0 ⟹ S = ( 0 2 0 ) + ( 1 2 0 ) i − ( 2 2 0 ) − ( 3 2 0 ) i + ( 4 2 0 ) − . . . + ( 2 0 2 0 ) = ℜ { ( 1 + i ) 2 0 } = ℜ { ( 2 i ) 1 0 } = ℜ { 2 1 0 ( − 1 ) } = − 1 0 2 4