Question
If a and d are two complex numbers, then the sum
to (n+1) terms of the following series
aC0 - (a + d)C1 + (a + 2d)C2 - ........... is
to (n+1) terms of the following series
aC0 - (a + d)C1 + (a + 2d)C2 - ........... is
Answer: Option C
:
C
We can write
aC0 - (a + d)C1 + (a + 2d)C2 - ........ upto (n+1) terms
=a(C0−C1+C2−........)+d(−C1+2C2−3C3+......) .............(i)
Again,(1−x)n=C0−C1x+C2x2−.........+(−1)nCnxn ..........(ii)
Differentiating with respect to x,
−n(1−x)n−1 = -C1 + 2C2x - .......... + (−1)nCnnxn−1 ......................(iii)
Putting x = 1 in (ii) and (iii), we get
C0 - C1 + C2 - ........ + (-1)nCn = 0
and -C1 + 2C2 - ........+(−1)nn.Cn = 0
Thus the required sum to (n+1) terms, by (i)
= a.0 + d.0 = 0.
Was this answer helpful ?
:
C
We can write
aC0 - (a + d)C1 + (a + 2d)C2 - ........ upto (n+1) terms
=a(C0−C1+C2−........)+d(−C1+2C2−3C3+......) .............(i)
Again,(1−x)n=C0−C1x+C2x2−.........+(−1)nCnxn ..........(ii)
Differentiating with respect to x,
−n(1−x)n−1 = -C1 + 2C2x - .......... + (−1)nCnnxn−1 ......................(iii)
Putting x = 1 in (ii) and (iii), we get
C0 - C1 + C2 - ........ + (-1)nCn = 0
and -C1 + 2C2 - ........+(−1)nn.Cn = 0
Thus the required sum to (n+1) terms, by (i)
= a.0 + d.0 = 0.
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution