IIT JEE , AIEEE Forum - Mathematics , Algebra - Some Progression and Series questions. please solve.

sahilgupta_iit

Q.1. Sum the series 1.1! + 2.2! + 3.3! + 4.4! + .......... to n terms.

Q.2. Show that (1 + 1/3).(1 + 1/3²).(1 + 1/3⁴).(1+ 1/3⁸)......(1 + 1/3^{2^n}) = 3/2 * (1 - 1/3^{2^(n+1)} ) .

Q.3. Show that (1/x+1) + (2/x² +1) + (4/x⁴ +1) + ........+ (2ⁿ/ x^{2^n} +1) = (1/x-1) - ( 2ⁿ⁺¹ / ^{x2^(n+1)} - 1 ).

Q.4. If n is a positive integer, prove that 2ⁿ > 1 + n*

2^n-1.

Greatdreams

1 at a time,

We have S =_{[x = 1 ]}

^[n]x(x!) = _{[x =1]}

^[n] {(x+1) - 1} (x!) = _[x=1]

^[n] (x+1)! - x!

= (n+1)! - 1

hsbhatt

Hints:

2. Let P = (1 + 1/3).(1 + 1/3²).(1 + 1/3⁴).(1+ 1/3⁸)......(1 + 1/3^{2^n}).

What is P*(1-1/3)?

3.Let S = (1/x+1) + (2/x² +1) + (4/x⁴ +1) + ........+ (2ⁿ/ x^{2^n} +1)

What is S - (1/x-1)?

Soln for 4:

2ⁿ-1 = 2^n-1+2^n-2+2^n-3+....+1

Now using AM-GM inequality,

(2^n-1+2^n-2+2^n-3+....+1)/n

ⁿ

(2^n-1* 2^n-2* 2^n-3*....*1) = ⁿ

2^{n-1 +n-2+..+1}

= ⁿ

2^(n-1)*n/2 =

2^n-1

Hence (2ⁿ-1)/n

2^n-1or 2ⁿ

1+n

2^n-1