Maths Binomial Theorem

T_r+1 =? C_r (3)^ (n-r)/2) (5)^ (r/8) (n ≥ r)
Clearly r should be a multiple of 8.
∴ there are exactly 33 integral terms
Possible values of r can be
0,8,16, . . .,32 * 8
∴ least value of n = 256

Kindly consider the following Image 

1. Let t? denotes r+1th term of (? x? +? x? )¹?
   t? = ¹? C? (? x? )¹? (? x? )? = ¹? C? ¹? x? ¹?
   If t? is independent of x
   90-15r=0? r=6
   This differs from the solution.
   Let's follow the solution's powers.
   (10-r)/9 - r/6 = 0? r=4
   maximum value of t? is 10K (given)
   ? ¹? C? is maximum
   By AM? GM (for positive numbers)
   (? ³/2+? ³/2+? ²/2+? ²/2)/4? (? /16)¹/?
   ? ? ? 16
   So, 10K = ¹? C?16
   ? K=336

Kindly consider the following figure

|x|<1, |y|<1, x? y
(x+y) + (x²+xy+y²) + (x³+x²y+xy²+y³) + .
By multiplying and dividing by x-y:
(x²-y²)+ (x³-y³)+ (x? -y? )+.)/ (x-y)
= (x²+x³+x? +.)- (y²+y³+y? +.)/ (x-y)
= (x²/ (1-x)- (y²/ (1-y)/ (x-y)
= (x²-y²-xy (x-y)/ (1-x) (1-y) (x-y)
= (x+y-xy)/ (1-x) (1-y)                                         

Sol. Let t? denotes r+1th term of (αx? + βx? )¹?
t? = ¹? C? (αx? )¹? (βx? )? = ¹? C? α¹? β? x? ¹?
If t? is independent of x
90-15r=0 ⇒ r=6
This differs from the solution.
Let's follow the solution's powers.
(10-r)/9 - r/6 = 0 ⇒ r=4
maximum value of t? is 10K (given)
⇒ ¹? C? α? β? is maximum
By AM ≥ GM (for positive numbers)
(α³/2+α³/2+β²/2+β²/2)/4 ≥ (α? β? /16)¹/?
⇒ α? β? ≤ 16
So, 10K = ¹? C?16
⇒ K=336