Q.1. The larger of 99^{50} + 100^{50} and 101^{50} is ................ (1982  2 Marks)
Ans. (101)^{50}
Sol. Consider (101)^{50} – {(99)^{50} + (100)^{50}} = (100 + 1)^{50} – (100 – 1)^{50} – (100)^{50}
= (100)^{50} [(1+ 0.0 1)^{50} – (1– 0.01)^{50} – 1]
= (100)^{50} [2 (^{50}C_{1}(0.01) + ^{50}C_{3}(0.01)^{3} + ....) – 1]
= (100)^{50} [2 (^{50}C_{3}(0.01)^{3} + ....)] > 0
∴ (101)^{50 }> (99)^{50} + (100)^{50}
∴ (101)^{50} is greater.
Q.2. The sum of the coefficients of the plynomial (1 + x – 3x^{2})^{2163} is ................ (1982  2 Marks)
Ans. 1
Sol. If we put x = 1 in the expansion of (1+ x – 3x2)^{2163} = A_{0} + A_{1}x + A_{2}x^{2} + ...
we will get the sum of coefficients of given polynomial, which clearly comes to be – 1.
Q.3. If (1 + ax )^{n} = 1 + 8x + 24x^{2} + .... then a = ...... and n = ............... (1983  2 Marks)
Ans. a = 2, n = 4
Sol. (1 + ax)^{n} = 1 + 8x + 24x^{2 }+ ...
⇒ (1 + ax)^{n} = 1 + nxa +
= 1 + 8 x + 24x 2+ ...
Comparing like powers of x we get
nax = 8x ⇒ na = 8 ....(1)
....(2)
Solving (1) and (2), n = 4, a = 2
Q.4. Let n be positive integer. If the coefficients of 2nd, 3rd, and 4th terms in the expansion of (1 + x)^{n} are in A.P., then the value of n is .............. (1994  2 Marks)
Ans. 7
Sol. We know that for a +ve integer n (1 + x)^{n} = ^{n}C_{0} +^{ n}C_{1 }x + ^{n}C_{2 }x^{2} + ......+ ^{n}C_{n }x^{n}
ATQ coefficients of 2^{nd}, 3^{nd}, and 4^{th} terms are in A.P. i.e.^{n}C_{1}, ^{n}C_{2}, ^{n}C_{3} are in A.P.
⇒ 2.^{n}C_{2} = ^{n}C_{1} + ^{n}C_{3}
⇒ n^{2} – 9n + 14 = 0
⇒ (n – 7) (n – 2) = 0 ⇒ n = 7 or 2
But for the existance of 4^{th} term, n = 7.
Q.5. The sum of the rational terms in th e expansion of is ................ (1997  2 Marks)
Ans. 41
Sol. Let T_{r +1} be the general term in the expansion of
Let T_{r +1}will be rational if 2^{5–r/2} and 3^{r/5} are rational numbers.
⇒ are integers.
⇒ r = 0 and r = 10 ⇒ T_{1} and T_{11} are rational terms.
⇒ Sum of T_{1} and T_{11} = ^{10}C_{0}25 – 0.30 + ^{10}C_{10}2^{5–5}.3^{2}
= 1.32.1 + 1.1.9 = 32 + 9 = 41
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