Find a and b if the sum of (ak+b)*k=8*[C(n+1,n-1)]^2/n? k=1 to k=n

First, we'll simplify the right side
term:

8*[C(n+1,n-1)]^2/n=
8*[C(n+1,n+1-n+1)]^2/n

We know that C(n+1,n-1) =
C(n+1,n+1-n+1) (they are complementary)

8*[C(n+1,n-1)]^2/n=
8*[C(n+1,2)]^2/n

C(n+1,2) =
n(n+1)/2

8*[C(n+1,n-1)]^2/n=
8*[n(n+1)/2]^2/n

8*[C(n+1,n-1)]^2/n=8*n^2*(n+1)^2/4n

We'll
simplify and we'll
get:

8*[C(n+1,n-1)]^2/n=2n(n+1)^2

Now,
we'll re-write the left term:

Sum(ak+b)*k= Sum(ak^2 +
bk)

Sum (ak+b)*k = a*Sum k^2 + b*Sum
k

Sum k^2 = 1^2 + 2^2 + ... +
n^2

Sum k^2 =
n(n+1)(2n+1)/6

Sum k = 1 + 2 + ... +
n

Sum k = n(n+1)/2

We'll
re-write the given identity:

a*n(n+1)(2n+1)/6 + b*n(n+1)/2
= 2n(n+1)^2

We'll multiply by six both
sides:

a*n(n+1)(2n+1) + 3*b*n(n+1) =
12n(n+1)^2

We'll factorize by
n(n+1):

n(n+1)[a(2n+1) + 3b] =
12n(n+1)^2

We'll divide by n(n+1) both
sides:

a(2n+1) + 3b = 12n +
12

We'll remove the
brackets:

2na + a + 3b = 12n +
12

Comparing, we'll get:

2na =
12n

We'll divide by 2n:

a =
6

a + 3b = 12

6 + 3b =
12

3b = 12-6

3b =
6

b = 2

The
values of a and b are: a = 6 and b = 2.