Binomial Theorem · Question 12

The sum of all possible values of n N, so that the coefficients of x, x2 and x3 in the expansion of (1 + x2)2(1 + x)n, are in arithmetic progression is :

1. 12
2. 9
3. 3
4. 7

The value of 100C5051 + 100C5152 + + 100C100101 is:

1. 2101101
2. 2100101
3. 2100100
4. 2101100

The coefficient of x48 in (1 + x) + 2(1 + x)2 + 3(1 + x)3 + + 100(1 + x)100 is equal to

1. 100 100C49 - 100C48
2. 100 101C49 - 101C50
3. 100C50 + 101C49
4. 100 100C49 - 100C50

Let Cr denote the coefficient of xr in the binomial expansion of (1 + x)n, n N, 0 r n. If Pn = C0 - C1 + 223C2 - 234C3 + + (-2)nn+1Cn, then the value of n=125 1P2n equals

1. 675
2. 580
3. 525
4. 650

If the coefficient of x in the expansion of (ax2 + bx + c)(1 - 2x)26 is -56 and the coefficients of x2 and x3 are both zero, then a + b + c is equal to :

1. 1483
2. 1300
3. 1500
4. 1403

If ( 115C0 + 115C1 ) ( 115C1 + 115C2 ) ( 115C12 + 115C13 ) = 1314C0 14C1 14C12, then 30 is equal to .

The number of integral terms in the expansion of (5 12 + 7 18 )1016 is:

1. 127
2. 128
3. 130
4. 129

The product of the last two digits of (1919)1919 is

The remainder when ((64)(64) )(64) is divided by 7 is equal to

1. 4
2. 6
3. 3
4. 1

The sum of the series 2 1 20C4 - 3 2 20C5 + 4 3 20C6 - 5 4 20C7 + + 18 17 20C20, is equal to .