Binomial Theorem

The expansion is (x + x^ (log? x)?
The (r+1)-th term is T? =? C? * x? * (x^ (log? x)?
The 4th term means r=3.
T? =? C? * x? * (x^ (log? x)³ = 35 * x? * x^ (3 log? x) = 35 * x^ (4 + 3 log? x).
Given T? = 4480.
35 * x^ (4 + 3 log? x) = 4480
x^ (4 + 3 log? x) = 4480 / 35 = 128.
x^ (4 + 3 log? x) = 128.
Take log? on both sides:
log? (x^ (4 + 3 log? x) = log? (128)
(4 + 3 log? x) * (log? x) = 7
Let t = log? x.
(4 + 3t)t = 7
3t² + 4t - 7 = 0
3t² - 3t + 7t - 7 = 0
3t (t-1) + 7 (t-1) = 0
(3t+7) (t-1) = 0
t = 1 or t = -7/3.
log? x = 1 ⇒ x = 2¹ = 2.
log? x = -7/3 ⇒ x = 2^ (-7/3).
Since x ∈ N, x = 2.

T? = ²²C? (x? )²²? x? ²?
T? = ²²C? x? ²²? ²?
22m - mr - 2r = 1
22m - 1 = r (m+2)
r = (22m-1)/ (m+2)
r = (22m+44-45)/ (m+2)
r = 22 - 45/ (m+2)
So possible value of m = 1,3,7,13,43
but ²? C? = 1540
only possible condition is m=13

N+5C_ {R-1}: N+5C_R: N+5C_ {R+1}
= 5:10:4
2 (N+5C_ {R-1}) = N+5C_R ⇒ 3R = N + 6
7 (N+5C_R) = 5 (N+5C_ {R+1}) ⇒ 12R = 18 + 5N
Solving: N = 6, R = 4
∴ the largest coefficient is N+5C_ {R+1} = 11C_5 = 462

Given (2x² + 3x + 4)¹? = Σ (r=0 to 20) a? x?
replace x by 2/x in above identity :-
2¹? (2x²+3x+4)¹? / x²? = Σ (r=0 to 20) a?2? /x?
⇒ 2¹? Σ (r=0 to 20) a? x? = Σ (r=0 to 20) a?2? x²? (from (i)
now, comparing coefficient of x? from both sides
(take r = 7 in L.H.S. and r = 13 in R.H.S.)
2¹? a? = a?2¹³ ⇒ a? /a? = 2³ = 8

Σ (r=30 to 50) 50-rC? =? C? +? C? +? C? + . + ³? C?
=? C? +? C? +? C? + . + (³? C? + ³? C? ) - ³? C?
=? C? +? C? +? C? + . + (³¹C? + ³¹C? ) - ³? C?
=? C? +? C? - ³? C?
=? ¹C? - ³? C?

T? =? C? (3x²/2)? (-1/3x)?
=? C? (3/2)? (-1/3)? x¹? ³? for the term independent of x put r = 6
=> T? =? C? (3/2)³ (-1/3)?
=? C? (1/6)³ = (9x8x7)/ (3x2x1) (1/6)³ = 7/18