# Given that k = 25 and f(k)=yk, calculate the sum y1+y2+...+yk f(x) = 3x + 12

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f(x) = 3x + 12

f(k) = 3k + 12 = yk

==> y1= 3*1 + 12 = 15

==> y2 = 3*2 + 12 = 18

==> y3= 3*3 + 12 = 21

==> Y4 =3*4+ 12 = 24

We notice that:

y1, y1, y3, ....yk is an A.P where y1= 15 and r= 3

........

==> yk = 15 + (k-1)*3

==> y25 = 15 + (24*3) = 87

Then,

S = 15 + 18 + 21+ ...+ 87

= 15 + (15+3) + (15+ 2*3) + ...+ (15+ 24*3)

= 15 + 15 + ...+ 3 + 2*3 + 3*3+ ...+ 24*3

= 15*25 + 3 ( 1+ 2+ ...+24)

= 375 + 3* (1+24)*24/2

= 375 + 3*25*12

= 375 + 900

= 1275

k =25, f(k) = yk. f(x) = 3x+12.

To find y1+y2+...yk.

Solution:

put k = n

yn = f(n) = 3n+12.

Therefore y(n+1) - yn = 3(n+1)+12- 3n-12 = 3n+3+12-3n-12 = 3 = d is the common difference of the AP.

y1 = 3*1+12 = 15 is th e 1st term of the A P

yn = 3*25+12 = 87 is the last term of AP

Therefore y1+y2+...yn = (y1+y25). d (n/2) = (15+87)*3*25/2 = 102*3*25/2 = 3825

First, we'll determine the terms of the sum, knowing that k = 25.

y1+y2+...+y25 = f(1)+f(2)+......+f(25)

To calculate the sum of values of f(x), we'll substitute x by the values 1, then 2, then 3, and so on, till 25, in the given expresison of the function.

f(1) = 3*1 + 12

f(2) = 3*2 + 12

f(3) = 3*3 + 12

.....................

f(25) = 3*25 + 12

We'll add the values above:

f(1) + f(2) + ...+f(25) = 3*(1+2+...+25) + 12*25

We notice that we can substitute the sum of the first 25 natural terms by the formula:

1+2+....+25 = (1+25)*25/2

f(1) + f(2) + ...+f(25) = 4*(26*25/2) + 300

f(1) + f(2) + ...+f(25) = 4*13*25+ 300

We'll factorize and we'll get:

f(1) + f(2) + ...+f(25) = 1300+300

**f(1) + f(2) + ...+f(25) = 1600**