If (1+x)n=C0+C1x+C2x2+.......Cnxn ....(i) then sum of seri

1.	If (1+x)n=C0+C1x+C2x2+.......Cnxn ....(i) then sum of series C0+Ck+C2k+....can be obtained by putting all roots of equation xk−1=0 in (i) & then adding vertically: for example: sum of C0+C2+C4.....can be obtained by putting all roots of equation x2=1 i.e. x=±1 in (i) At x=1 C0+C1+C2..........Cn=2n x=−1 C0−C1+C2−C3...........=0 Adding we get C0+C2+C4.....=2n−1 Now answer the folloiwng If n is a multiple of 3, then C0+C3+C6+................ equals
Answer» $I f (1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . C_{n} x^{n} . . . . (i)$ then sum of series $C_{0} + C_{k} + C_{2 k} +$ ....can be obtained by putting all roots of equation $x^{k} - 1 = 0$ in (i) & then adding vertically: for example: sum of $C_{0} + C_{2} + C_{4}$ .....can be obtained by putting all roots of equation $x^{2} = 1$ i.e. $x = \pm 1$ in (i) At $x = 1 C_{0} + C_{1} + C_{2} . . . . . . . . . . C_{n} = 2^{n}$ $x = - 1 C_{0} - C_{1} + C_{2} - C_{3} . . . . . . . . . . . = 0$ Adding we get $C_{0} + C_{2} + C_{4} . . . . . = 2^{n - 1}$ Now answer the folloiwng If n is a multiple of 3, then $C_{0} + C_{3} + C_{6} + . . . . . . . . . . . . . . . .$ equals

If (1+x)n=C0+C1x+C2x2+.......Cnxn ....(i) then sum of series C0+Ck+C2k+....can be obtained by putting all roots of equation xk−1=0 in (i) & then adding vertically: for example: sum of C0+C2+C4.....can be obtained by putting all roots of equation x2=1 i.e. x=±1 in (i) At x=1 C0+C1+C2..........Cn=2n x=−1 C0−C1+C2−C3...........=0 Adding we get C0+C2+C4.....=2n−1 Now answer the folloiwng If n is a multiple of 3, then C0+C3+C6+................ equals

Answer»

$I f (1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . C_{n} x^{n} . . . . (i)$
then sum of series $C_{0} + C_{k} + C_{2 k} +$ ....can be obtained by putting all roots of equation $x^{k} - 1 = 0$ in (i) & then adding vertically:
for example: sum of $C_{0} + C_{2} + C_{4}$ .....can be obtained by putting all roots of equation $x^{2} = 1$
i.e. $x = \pm 1$ in (i)
At $x = 1 C_{0} + C_{1} + C_{2} . . . . . . . . . . C_{n} = 2^{n}$
$x = - 1 C_{0} - C_{1} + C_{2} - C_{3} . . . . . . . . . . . = 0$
Adding we get $C_{0} + C_{2} + C_{4} . . . . . = 2^{n - 1}$

Now answer the folloiwng

If n is a multiple of 3, then $C_{0} + C_{3} + C_{6} + . . . . . . . . . . . . . . . .$ equals

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