If the coefficients of rth, (r+1)th, and (r + 2)th terms in the binomial expansion of $${{\rm{(1 + y )}}^m}$$ are in A.P., then m and r satisfy the equation
In option C, $${m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0$$ putting r = 2 we get
m2 - 9m + 14 = 0. So Option C is correct.
2
AIEEE 2005
MCQ (Single Correct Answer)
If $$x$$ is so small that $${x^3}$$ and higher powers of $$x$$ may be neglected, then $${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$$ may be approximated as
[ As x3 is so small we can ignore $$-{3 \over {16}}{x^3}$$]
= $$ - {3 \over 8}{x^2}$$
3
AIEEE 2005
MCQ (Single Correct Answer)
If the coefficient of $${x^7}$$ in $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ equals the coefficient of $${x^{ - 7}}$$ in $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$, then $$a$$ and $$b$$ satisfy the relation
A
$$a - b = 1$$
B
$$a + b = 1$$
C
$${a \over b} = 1$$
D
$$ab = 1$$
Explanation
General term of $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ is Tr+1.
= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( { - 1} \right)^r}{\left( b \right)^{ - r}}{\left( x \right)^{11 - r}}{\left( x \right)^{ - 2r}}$$
For the coefficient of x-7,
11 - 3r = -7
$$ \Rightarrow $$ r = 6
$$\therefore$$ Coefficient of x-7 = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$
According to question,
Coefficient of x7 = Coefficient of x-7
$$ \Rightarrow $$ $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$
$$ \Rightarrow $$ $$ab = 1$$
4
AIEEE 2005
MCQ (Single Correct Answer)
If $$A = \left[ {\matrix{
1 & 0 \cr
1 & 1 \cr
} } \right]$$ and $$I = \left[ {\matrix{
1 & 0 \cr
0 & 1 \cr
} } \right],$$ then which one of the following holds for all $$n \ge 1,$$ by the principle of mathematical induction