sum of series?

Answer: √8 = 2√2.

Solution:

i) This is with the application of (1 - x)^(-p/q)

ii) (1 - x)^(-p/q) =

= 1 + (-p/q)(-x) + {(-p/q)(-p/q - 1)/2!}*(-x)² + {(-p/q)(-p/q - 1)(-p/q - 2)/3!}*(-3)³ + .....

= 1 + (p/1!)*(x/q) + {p(p + q)/2!}*(x/q)² + {p(p + q)(p + 2q)/3!}*(x/q)³ + ......

iii) Here, S = 1 + 3/4 + (3*5)/(4*8) + (3*5*7)/(4*8*12) + .......

==> S = 1 + 3*(1/4) + {(3*5)/(1*2)}*(1/4)² + {(3*5*7)/(1*2*3)}*(1/4)³ + ...

==> S =1 + (3/1!)*{(1/2)/2} + {3*(3 + 2)/2!}*{(1/2)/2}² + {3*(3+2)*(3 + 2*2)/3!}*{(1/2)/2}³ + .

So here p = 3; q = 2 and x = 1/2

Applying the formula given above, this S = (1 - 1/2)^(-3/2) = (1/2)^(-3/2) = 2^(3/2)

Thus answer = √8 = 2√2

1+ 3/4 + (3.5)/(4.8) + (3.5.7)/(4.8.12) +... is less than 4 ANSWER

as 1+ 3/4 + (3.5)/(4.8) + (3.5.7)/(4.8.12) +... < 1 + 3/4 + (3/4)^2 + (3/4)^3 + ...... = 1 / [1 -- 3/4] = 4