2011/059) For all n>= 1 , prove that:- (1)+(1/1+2) + (1/1+2+3)+ ....... + (1/1+2+3+......+n)=(2n/n+1)

this type of sum come under telescopic sum

k th term = 1/(1+2 +3 + ... k) = 2/(k(k+i) = 2 * ( 1/k - 1/(k+1))

1st term = 2(1/1 - 1/2)
2nd term = 2(1/2 - 1/3)

nth term = 2( 1/n - 1/(n+1))

sum (all terms except 1st and last cancel) = 2/(1/1- 1/(n+1)) = 2n/(n+1)

2011/058)find the highest power of 2 that divides a number from n to m

for example from 12 to 14 the highest power of 2 that divides is 2^2 that divides 12.

this can be done by factoring all numbers from n to m in form 2^k * l and then finding the highest k however that can be a very slow process. and this can be improved signifcantly

let n = 2^p * q where q is odd( p = 0 in case n is odd)

next multiple of 2^p shall be n + 2^p

so we have algorithm

let n = 2 ^p* q

set r = 2 ^ p + q (1st number)
loop step:
if r + 2^p > m exit
add 2 ^p to r and 1 to q( r = r + 2^p , q = q + 1)
keep deviding q by 2 adding 1 to p until q is odd
go to loop step:
r is the value and p is the highest power

suppose we need to find between 21 and 31

21 = 2^0 * 21
add 1 22 = 2 ^ 0 * 22 or 2^1 * 11
add 2^1 or 2, 24 = 2^1 * 12 or 2^3 *3
add 2^3 or 8 we get 32 > 31

so and = 24 = 2^3 * 3

2011/057) Given |a | < 1 and |b| < 1

If $1+a+a^2+\cdots =x$ and $1+b+b^2+\cdots.=y$ then find $1+ab+a^2.b^2+.... $ in terms of x and y

Solution
we have
$x = \dfrac{1}{1-a}$
or $(1-a) = \dfrac{1}{x}$
or $a = 1 – \dfrac{1}{x} = \dfrac{x-1}{x}$
similarly
$b = \dfrac{y-1}{y}$
now as |a | and |b| both < 1 so |ab| < 1 and hence
$1+ ab + a^2b^2 + a^3b^3 + \cdots= \dfrac{1}{1-ab} = \dfrac{1}{1- \frac{x-1}{x} * \frac{y-1}{y}}$
= $\dfrac{xy}{xy - (x - 1)(y - 1)}$
= $\dfrac{xy}{x + y - 1}$ 

2011/056) Find lim n-> inf (n/(n+2))^n

(n/(n+2))^n
= (1/(1+2/n)^n
= ((1/(1+2/n)^n/2)^2

As (1+1/x) ^x as x -> infinite is e so /(1+2/n)^n/2 = e so given limit = 1/e^2

2011/055) Find the sum of rational terms of (2^(1/2) + 3^(1/5))^ 10

Let a rational term be (10Cn) (2^1/2)^n (3(^1/5)^(10-n)

Now n/2 and (10-n)/5 need to be integers

n = 0 and n =1 10 satisfy the criteria and for n = 0 we get 9 and for n = 10 we get 32 so sum of rational terms = 41

2011/054) If a = e^(2ipi/7) and f(x) = A0 + sum ( k = 1 to 20) Ak x^k find the value of f(a) + f(ax) + f(a^2x) + f(a^3x) + + f(a^4x)+ + f(a^5x) + f(a^6x)

We have a = e^(2ipi/7)

So a^7 = e^(2ipi) = 1

or a^(7 t) = 1 or

or (a^7t) – 1 = 0

or (a^t-1)(a^t + a^2t + a^3t + a^4 t + a^5 t + a^6 t) = 0

if t is multiple of 7 then a^k = 1

and if t is not multiple of 7 then a^t-1 is not zero so a^t + a^2t + a^3t + a^4 t + a^5 t + a^6 t) = 0

so in f(a) + f(ax) + f(a^2x) + f(a^3x) + + f(a^4x)+ + f(a^5x) + f(a^6x)

for k multiple of 7 sum is 7 and for k not multiple of 7 sum of x^k = 0

so sum = 7 A0 + 7A7 x^7 + 7 A14 x^14