IIT JEE

If |x| < 1 and |y| < 1 then prove that

(x +y) +(x^{2}
+ xy + y^{2}) + (x^{3} +x^{2}y+xy^{2} + y^{3})
+ . .. .¥ = .

The given sum S =
(x + y) +(x^{2} + xy + y^{2}) + . . . . .

=

=

= == .

Let S = and show that
assuming
not all a_{i} ‘ s are equal .

AM_{1}
= GM_{1}
= (P denotes product sign)

and AM_{1}
> GM_{1} (as a_{i} ‘s are not all equal) . . . .
(1) .

Again
AM_{2} = and
GM_{2} =

and
AM_{2} > GM_{2} ..
. . . (2).

By multiplying (1) and (2)

> 1

&⇒

&⇒ .

If n is a root of the
equation x^{2}(1 – ac) – x(a^{2} + c^{2}) – (1 + ac) =
0 and if n harmonic means are inserted between a and c show that the
difference between the first and last mean is equal to ac(a –c). .

Given: n^{2} (1 –ac) – n(a^{2}
+c^{2}) –(1 + ac) = 0 . . . (1).

also, a, H_{1} H_{2} H_{3}
……..H_{n} c are in H.P.

To prove H_{1} –H_{n} =
ac(a –c)

Now are in A.P.

&⇒ or d =

Find the sum to n terms of the series

The rth terms of the series is given by

t_{r} =

&⇒ t_{r + 1} =

&⇒ rt_{r} = (r + 4)t_{r +1}

&⇒ rt_{r} - (r
+ 1)t_{r + 1} = 3t_{r + 1}

Putting r = 1, 2, ...., n - 1 we get.

1t_{1} - 2t_{2}
= 3t_{2}

2t_{2} - 3t_{3} = 3t_{3}

. . .

. . .

. . .

(n - 1)t_{n - 1}
- nt_{n} = 3t_{n}

Adding the above equations, we get

t_{1} - nt_{n}
= 3[t_{2} + t_{3} + .... + t_{n}] .

&⇒ 4t_{1} - nt_{n} =
3[t_{1} + t_{2} + .... + t_{n}].

&⇒

&⇒ t_{1} + t_{2} +
..... + t_{n} = - .

Let a, b, c be three distinct
positive real numbers in G.P., then

prove that a^{2} + 2bc – 3ac > 0 . .

Since
a, b, c ∈ R^{+} and
distinct

&⇒ AM > GM > HM

since b = and consider AM and HM of a and c,

&⇒

From first inequality (a +c) > 2b

&⇒ a^{2} +ac – 2ab > 0

From second inequality b(a +c) > 2ac

&⇒ 2ab + 2bc –4ac > 0

Adding the two
inequalities a^{2} +2bc –3ac > 0 . .