t is an number in the series: 1/t + 3/t + 5/t + ...+ t-1/t a)Determine the number of terms in the series in terms of t. b)Determine the sum of the series in terms of t. c)Hence, or otherwise, evaluate:(1/4+3/4)+(1/6+3/6+5/6)+(1 /8+3/8+5/8+7/8)+...+(1/50+3/50+5/50 +49/50)​

Below in bold.

Step-by-step explanation:

1/t + 3/t + 5/t +  ...+  t-1/t

This is an arithmetic series in which the common difference is 2/t.

a) The number of terms = [ (last term - first term) / common difference] + 1

=  [(t-1)/t - 1/t) / 2/t]  + 1

= (t - 2)/t * t/2 + 1

= (t -2)/2 + 1

= (t-2) / 2 + 2/2

= t/2.

b). Sum of the series (n/2)(a + l)  where a = first term , n = number of terms and l = last term, so

Sum = (t/2)/2 ( 1/t + (t-1)/t)

= t/4 * 1

= t/4.

c). (1/4+3/4 )+ (1/6+3/6+5/6 )+( 1/8+3/8+5/8+7/8) +...+(1/50+3/50+5/50+49/50)​

Using the results from a) and b):

The sum = 4/4  + 6/4 + 8/4 + + 50/4.

= 1 + 1.5 + 2 + + 12.5

This is an AS  with common difference = 0.5,

Number of terms = [(12.5 -1) / 0.5] + 1

= 23 + 1

= 24, so

Sum = (24/2)(1 + 12.5)

=  12  * 13.5

=  162.

where is the figure

u bout ugly as   a mf

step-by-step explanation: lol