Let `S_n=1+q+q^2 +?+q^n` and `T_n =1+((q+1)/2)+((q+1)/2)^2+?+((q+1)/2)^n` If `alpha T_100=^101C_1 +^101C_2 xS_1 +^101C_101 xS_100,` then the value of `alpha` is equal to (A) `2^99` (B) `2^101` (C) `2^100` (D) `-2^100`

Options

A. `2^(100)`
B. 202
C. 200
D. `2^(99)`

Answer

A

Explanation

(a) We have, `S_(n) = 1 + q + q^(2) + ...+ q^(n)` and `T_(n) = 1 + ((q +1)/(2)) + ((q +1)/(2))^(2) + ...+ ((q +1)/(2))^(n)`
Also, we have
`.^(101)C_(1) + .^(101)C_(2) + .^(101)C_(3)S_(2) + ...+ .^(101)C_(101) S_(100) = alpha T_(100)`
`rArr .^(101)C_(1) + .^(101)C_(2) ( 1 + q)+ .^(101)C_(3) (1 + q + q^(2)) + ...+ .^(101)C_(101) (1 + q + q^(2) + ... + q^(100))`
`= alpha. T_(100)`
`rArr .^(101)C_(1) + .^(101)C_(2) ((1 -a^(2)))/(1 -q) + .^(101)C_(3) ((1 -q^(3))/(1 -q)) + .^(101)C_(4) ((1 -q^(4))/(1- q)) + ...+ .^(101)C_(101) ((1 -q^(101))/(1 -q))`
`= alpha.T_(100) [ :' " for a GP", S_(n) = a ((1 -r^(n))/(1 -r)), r != 1]`
`rArr (1)/(1 -q) [{.^(101)C_(1) + .^(101)C_(2) +...+ .^(101)C_(101)} - {.^(101)C_(1) q + .^(101)C_(2)q^(2) + ...+ .^(101)C_(101) q^(101)} = alpha.T_(100)`
`rArr (1)/((1 -q)) [(2^(101) -1) - ((1 + q)^(101) -1)] = alpha T_(100)`
`rArr (2^(101) - (q +1)^(101))/(1 -q) = alpha [1 + (q +1)/(2) + ((q +1)/(2))^(2) + ...+ ((q +1)/(2))^(100)]`
`rArr (2^(101) - (q + 1)^(101))/(1 -q) = alpha [1.(1 - ((q +1)/(2))^(101))/(1 - (q +1)/(2))] " " [ :' q != 1 rArr q + 1 != 2 rArr (q +1)/(2) != 1]`
`= (alpha [2^(101) - (q + 1)^(101)])/((1 -q).2^(100)) rArr alpha = 2^(100)`