Binomial TheoremHard
Question
The sum of the coefficients of all the integral powers of x in the expansion of (1 + 2√x)40 is :
Options
A.340 + 1
B.340 - 1
C.
(340 - 1)
(340 - 1)D.(340 + 1)
(340 + 1)Solution
(1 + 2√x)40 = 40C0 + 40C1 2√x + .....+ 40C40 (2√x)40
(1 - 2√x)40 = 40C0 - 40C1 2√x + ...... + 40C40 (2√x)40
(1 + 2√x)40 + (1 - 2√x)40
= 2[40C0 + 40C2 (2√x)2 + ......+ 40C40 (2√x)40]
Putting x = 1
40C0 + 40C2(2)2 +.......+ 40C40 (2)40 =
(1 - 2√x)40 = 40C0 - 40C1 2√x + ...... + 40C40 (2√x)40
(1 + 2√x)40 + (1 - 2√x)40
= 2[40C0 + 40C2 (2√x)2 + ......+ 40C40 (2√x)40]
Putting x = 1
40C0 + 40C2(2)2 +.......+ 40C40 (2)40 =
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