The sum of the coefficients of (6a - 5b)n, where n is a positive integer, is
1
- 1
2n
2n - 1
A.
1
For the sum of the coefficients in the expansion of (6a - 5b)n, put a = b = 1
If x is numerically so small so that x2 and higher power of x can be neglected, then is approximately equal to
If S1, S2 and S3 are the sums of n, 2n and 3n terms of an arithmetic progression respectively, then
S2 = 3S3 - 2S1
S3 = 4(S1 + S2)
S3 = 3(S2 - S1)
S3 = 2(S2 + S1)