Test: Binomial Theorem - 6

Since a, b, c are in A.P., therefore  

z_r are the roots of z¹⁰¹ − 1 =0 , except 1 

For the sum of coefficient, put x = 1, to obtain the sum is (1 + 1 ‐ 3)²¹³⁴ = 1

C(n, r) + c(n ‐1, r) + C(n ‐ 2, r) + . . . + C(r, r)

= ⁿ⁺¹C_r+1 (applying same rule again and again)
(∵ ⁿC_r + ⁿC_r‐1 = ⁿ⁺¹C_r)

Here last term is of 14 degree.

The general term 
The term independent of x, (or the constant term) corresponds to x^18−3r being x⁰ or 18 − 3r= 0 ⇒ r = 6 .

Hence, t₈ is the greatest term and its value is

∴ 
For first integral term for r = 3;

(2^1/5 + 3^1/10)⁵⁵
Total terms = 55 + 1 = 56

Here r = 0, 10, 20, 30, 40, 50
Number of rational terms = 6;
Number of irrational terms = 56 ‐ 6 = 50

We have, (2x + 3y − 4z)ⁿ = {2x + (3y − 4z)}ⁿ


Clearly, the first term in the above expansion gives one term, second term gives two terms, third term gives three terms and so on.
So, Total number of term = 1 +2+3+...+n+(n+1)

Given expansion is 

Since, we have to find coefficient of x⁻¹⁰
∴ −12 + 2r = −10 ⇒ r = 1
Now, then coefficient of x⁻¹⁰ is ¹²C₁(a)¹¹(b)¹ = 12a b

 it has 12 terms in it’s expansion ,
so there are two middle terms (6th and 7th);

Here 2n is even integer, therefore, term will be the middle term.

Given sum = sum of odd terms