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

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### 1 Answer

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.**