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`        (2n+1C0)2 - (2n+1C1)2 + (2n+1C2) – …... – (2n+1C2n+1) = ?Please answer the above series `
2 years ago

```							Seeing  the question it is clear that  ans will be of the form (a-b) ^2n+1 Expanding  (a-b) ^2n+1a^2n+1- 2n+1(a^2n)b + n ( 2n+1)a^2n-1 b^2_ _ _ _ - b^2n+1          __1Now from(1) and the given equation a^2n+1=1                       __22n+1(a^2n)b=(2n+1)^2= a^2n b = 2n +1          __3n(2n+1)a^2n-1 b^2 = n(2n +1)= a^2n-1 b^2 =1          __4b^2n+1 =1                   __5Now dividing  4/3a^2n b/a^2n-1 b^2 = 2n + 1= a/b = 2n + 1              __6Now solving (2)&(5)a^2n+1 = b^2n+1=(a/b)^2n+1 = 1=(2n+1)^2n+1 = 1  (by 6)Let 2n+1 = x    __7Then,  x^x = 1= x=0,  or  x= 1By 72n+1 = 0,  or  2n+1 =1Now as 2n+1 is odd,  it is never equal to 0=> 2n+1 =1   __8Using the value in equation  (2)&(5),We get,  a=1          &   b=1Now,  (a-b) ^2n+1          = (1-1)^2(0)+1=0^1 =0 Answer = 0
```
2 years ago
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