Prove that `n C_0+^n C_3+^n C_6+=1/3(2^n+2cos(npi)/3)` .

Prove that ^n C_0 ^(2n)C_n- ^n C_1 ^(2n-2)C_n+ ^n C_2^(2n-4)C_n-=2^ndot

Prove that (^n C_0)/1+(^n C_2)/3+(^n C_4)/5+(^n C_6)/7+.....+dot=(2^n)/(n+1)dot

Prove that ^n C_1(^n C_2)(^n C_3)^3(^n C_n)^nlt=((2^n)/(n+1))^(n+1_C()_2),AAn in Ndot

Prove that "^n C_0^(2n)C_n-^n C_1^(2n-1)C_n+^n C_2xx^(2n-2)C_n++(-1)^n^n C_n^n C_n=1.

Prove that .^(n)C_(0) - ^(n)C_(1) + .^(n)C_(2)- ^(n)C_(3) + "…" + (-1)^(r) + .^(n)C_(r) + "…" = (-1)^(r ) xx .^(n-1)C_(r ) .

Prove that ^n C_0^n C_0-^(n+1)C_1^n C_1+^(n+2)C_2^n C_2-=(-1)^ndot

Prove that (a) (1+i)^n+(1-i)^n=2^((n+2)/2).cos((npi)/4) , where n is a positive integer. (b) (1+isqrt(3))^n+(1-isqrt(3)^n=2^(n+1)cos((npi)/3) , where n is a positive integer

Prove that (r+1)^n C_r-r^n C_r+(r-1)^n C_2-^n C_3++(-1)^r^n C_r = (-1)^r^(n-2)C_rdot

Using binomial theorem (without using the formula for ^n C_r ) , prove that "^n C_4+^m C_2-^m C_1^n C_2 = ^m C_4-^(m+n)C_1^m C_3+^(m+n)C_2^m C_2-^(m+n)C_3^m C_1+^ (m+n)C_4dot

Prove that .^(n)C_(1) + 2 .^(n)C_(2) + 3 .^(n)C_(3) + "…." + n . ^(n)C_(n) = n 2^(n-1) .