`5C_0+8.C_1+11.C_2+....` to `(n+1)` terms =

C_0+4. C_1 + 7. C_2+…...(n+1) terms =

3. C_0 + 7. C_1 + 11. C_2 +…...+ (4n+3) . C_n=

3. C_0 + 7. C_1 + 11. C_2 +…...+ (4n+3) . C_n=

If (1+x)^n=C_0+C_1x+C_2x^2+……..+C_nx^n in N prove that (a) 3 C_0- 8C_1+13C_2-18C_3+...."upto" (n+1) term=0 if n ge 2 (b ) 2C_0+2^2(C_1)/(2)+2^3(C_2)/(3)+2^4C_(3)/4+....+2^(n+1)(C_n)/(n+1)=(3^n+1-1)/(n+1) ( c) C_0^2+(C_1^2)/2+C_2^2/3+....+C_n^2/(n+1)=((2n+1)!)/(((n+1)!)^2)

Show that 3C_0-8C_1 + 13C_2 - 18C_3 + ..... + (n+1)^(th) term = 0

C_0 C_1 + C_1 C_2 + .... + C_(n-1) C_n = (2^n.n.1.3.5... (2n-1))/(n+1)

lf C_(r)=""^(n)C_(r) , then C_(0)-1/3C_(1)+1/5C_(2) …… upto (n+1) terms equal

The value of C_(0)^(2)+3.C_(1)^(2)+5.C_(2)^(2)+..... to (n+1) terms where C_(r)=^(n)C_(r ) is