Use mathematical induction to prove that

Question

8+11+14+17+... + (3n+5) = n/2 (3n+13)

Answer 1

Add (3(n+1)+5) to both sides

then try to reduce the right side to the form (n+1)/2(3(n+1)+13)

transform n/2(3n+13) + (3(n+1)+5) into (n+1)/2(3(n+1)+13

first show it's true for n=1

as the 1st term is 8, and (3(1)+5) = 8 and 1/2(3+13) = 16/2 = 8

then show it's true for n=2

as the 2nd term is 11 and 2/2(3x2+13) = (19) = 8 + 11, the sum of the first two terms

then assume it is true for (3n+5), for all n, and that that leads to its being true for (3(n+1)+5)

That's the tricky part. requiring some algebra and manipulations

but you can see it's true, let n=3

left side is 8 + 11 + (3n+5) = 8+11+14 = 33

for n=4

8+11+14+17 = 33+17 = 50

the right side for n=4 is 4/2 (3x4+13) = 2(12+13) = 2(25)=50

Answer 2

8+11+14+17+...+(3n+5)=(n/2)(3n+13)

show it's true for n=1, which it is; 8=8

assume the original equation is true

show that the equation is true for "n+1" (replace each "n" by "n+1")

8+11+14+17+...+3(n+1)+5=[(n+1)/2][3(n+1)+13]

8+11+14+17+ +3n+3+5=[(n+1)/2][3n+3+13]

8+11+14+17+...+(3n+8)=[(n+1)/2][3n+16]

8+11+14+17+...+(3n+8)=(n+1)(3n+16)/2

8+11+14+17+ +(3n+5)+(3n+8)=(3n^2+19n+16)/2

8+11+14+17+...+(3n+5)+(3n+8)=(3n^2/2)+(19n/2)+8

now to verify this you add (3n+8) to the original right side of the original equation to see if you get this new right side because (3n+8) was added to the original left side of the original equation (don't forget that the term that precedes (3n+8) is the last term of the original equation which is (3n+5))

(n/2)(3n+13)+(3n+8)=

(3n^2/2)+(13n/2)+3n+8=

(3n^2/2)+(13n/2)+(6n/2)+8=

(3n^2/2)+(19n/2)+8 which is exactly the new right side of the new equation

thus it is proved by mathematical induction