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Prove that following i) 2.C(0)+5.C(1)+...

Prove that following
i) `2.C_(0)+5.C_(1)+8.C_(2)+…..+(3n+2)C_(n)=(3n+4).2^(n-1)`

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With usual notations prove that 2.C_0 + 7.C_1 + 12.C_2 + ……..+(5n + 2).C_n = (5n + 4).2^(n-1)

If C_(0),C_(1),C_(2)…….,C_(n) are the combinatorial coefficient in the expansion of (1+x)^n, n, ne N , then prove that following C_(1)+2C_(2)+3C_(3)+..+n.C_(n)=n.2^(n-1) C_(0)+2C_(1)+3C_(2)+......+(n+1)C_(n)=(n+2)C_(n)=(n+2)2^(n-1) C_(0),+3C_(1)+5C_(2)+.....+(2n+1)C_n =(n+1)2^n (C_0+C_1)(C_1+C_2)(C_2+C_3)......(C_(n-1)+C_n)=(C_0.C_1.C_2....C_(n-1)(n+1)^n)/(n!) 1.C_0^2+3.C_1^2+....+ (2n+1)C_n^2=((n+1)(2n)!)/(n! n!)

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If (1+x)^(n)=C_(0)+C_(1).x+C_(2).x^(2)+….+C_(n).x^(n). then prove that (i) C_(0)+2C_(1)+3C_(2)+…+(n-1)C_(n)=(n+2).2^(n-1) (ii)C_(0)+3C_(1)+5C_(2)+...+(2n+1)C_(n)=(n+1).2^(n)

Prove that C_3 + 2.C_4+ 3.C_5 + ……..+ (n-2).C_n = (n-4).2^(n-1) + (n+2) where n > 3

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Find the following sums : (i) .^(n)C_(0)-.^(n)C_(2)+.^(n)C_(4)-.^(n)C_(6)+"....." (ii) .^(n)C_(1)-.^(n)C_(3)+.^(n)C_(5)-.^(n)C_(7)+"...." (iii) .^(n)C_(0)+.^(n)C_(4)+.^(n)C_(8)+.^(n)C_(12)+"....." (iv) .^(n)C_(2) + .^(n)C_(6) + .^(n)C_(10)+.^(n)C_(14)+"......" (v) .^(n)C_(1) + .^(n)C_(5)+.^(n)C_(9)+.^(n)C_(13)+"...." (vi) .^(n)C_(3) + .^(n)C_(7) + .^(n)C_(11) + .^(n)C_(15) + "....."