Binomial Theorem Part 3

Let f(x)=(1+x+x2)10=

20

∑

r=0

arxr
The sum of odd coefficients: S‌odd ‌=a1+a3+a5+...+a19
Subtracting 11a2 from above will give the answer
‌S‌odd ‌=‌

f(1)−f(−1)

2

‌f(1)=(1+1+1)10=310
‌f(−1)=(1−1+1)10=(1)10=1
‌S‌odd ‌=∑‌odd ‌rar=‌

310−1

2

Now for a2

Now use:
‌(1−x3)10=

10

∑

x=0

(−1)k(

10

k

)x3k
‌(1−x)−10=

∞

∑

r=0

(

r+9

9

)xr
So
f(x)=(

10

∑

k=0

(−1)k(

10

k

)x3k)⋅(

∞

∑

r=0

(

r+9

9

)xr)
Only the term with x0 from the first sum (i.e., k=0 ) can contribute to x2, since all other k≥1 gives x3k≥x3
From (1−x3)10 : the x0 term is (

10

0

)=1

From (1−x)−10 : the coefficient of x2 is
(

2+9

9

)=(

11

9

)=55
Hence, a2=1.55=55
Now, S‌odd ‌−11a2=‌

310−1

2

−11⋅55=121k
310=59049
So:
‌S=‌

59049−1

2

−605=‌

59048

2

−605
‌=29524−605=28919
So:121k=28919⇒k=‌

28919

121

=239