Grade 11Algebra
Plz answer fast,..........,...,......,........................
Plz answer fast,..........,...,......,........................
1 Answer
Rajat
Summation={(4n+1)C(0) + (4n+1)C(2n)} + .......{(4n+1)C(n-1)+(4n+1)C(2n-n+1)}+{(4n+1)C(n)+ (4n+1)C(2n-n)}
= {(4n+1)C(0)+(4n+1)C(1)+......+(4n+1)C(n)+......+(4n+1)C(2n)} +(4n+1)C(n)
=(1/2){(4n+1)C(0)+.....(4n+1)C(2n)....(4n+1)C(4n+1)}+(4n+1)C(n) . ........(1)
Since, [[(X)C(Y)=(X)C(X-Y)]]
From 1,
Assume the binomial expansion of (1+1)(4n+1) =2(4n+1)
So required summation = (1/2)2(4n+1) + (4n+1)C(n)
=24n+(4n+1)C(n)
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