Q Prove that ∑r=1k -3r-1 3nC2r -1 = 0, where K =3n2 and n is an even - Maths - Binomial Theorem

Question

Q Prove that ∑ r = 1 k   - 3 r - 1   3 n C 2 r
  - 1     =   0 ,   w h e r e   K
  = 3 n 2   and n is an even positive integer

Lovina Kansal · Accepted Answer

Dear student
Since n is even integer
Let n=2mk=3n2=32m2=3mThe summation becomesS=∑r=1k-3r-1 3nC2r-1=∑r=13m-3r-1 6mC2r-1   ∵n=2m and k=3m
Now,  1+x6m=6mC0+6mC1x+6mC2x2+
+6mC6m-1x6m-1+6mC6m x6m     (1)
and 1-x6m=6mC0+6mC1-x+6mC2-x2+
+6mC6m-1-x6m-1+6mC6m -x6m     (2) So,
1+x6m-1-x6m=6mC0+6mC1x+6mC2x2+
+6mC6m-1x6m-1+6mC6m x6m-6mC0+6mC1-x+6mC2-x2+
+6mC6m-1-x6m-1+6mC6m -x6m
1+x6m-1-x6m⇒1+x6m-1-x6m=26mC1x+6mC3x3+ +6mC6m-1x6m-1
⇒1+x6m-1-x6m2x=6mC1+6mC3x2+ +6mC6m-1x6m-2
Let x2=y⇒x=y 1+y6m-1-y6m2y=6mC1+6mC3y+ +6mC6m-1y3m-1
With y=-3, RHS becomes equals to S
LHS=1+i36m-1-i36m23i
Let z=1+3i=2eπ3iand z¯=1-3i=2e-π3
∴S=z6m-z¯6m2i3=2isin6m×π22i3=sin2mπ2i3=0
Hence proved
Regards

Q. Prove that ∑ r = 1 k - 3 r - 1 3 n C 2 r - 1 = 0 , w h e r e K = 3 n 2 and n is an even positive integer.

Q. Prove that $\sum_{r = 1}^{k} {(- 3)}^{r - 1 3 n} C_{2 r - 1} = 0, w h e r e K = \frac{3 n}{2}$ and n is an even positive integer.