Answer:
The sum of the first 650 terms of the given arithmetic sequence is 2,322,775
Step-by-step explanation:
The first term here is 4
while the nth term would be ai = a(i-1) + 11
Kindly note that i and 1 are subscript of a
Mathematically, the sum of n terms of an arithmetic sequence can be calculated using the formula
Sn = n/2[2a + (n-1)d)
Here, our n is 650, a is 4, d is the difference between two successive terms which is 11.
Plugging these values, we have
Sn = (650/2) (2(4) + (650-1)11)
Sn = 325(8 + 7,139)
Sn = 325(7,147)
Sn = 2,322,775
Let
Differentiating twice gives
When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.
Substitute these into the given differential equation:
Then the coefficients in the power series solution are governed by the recurrence relation,
Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.
• If n is even, then n = 2k for some integer k ≥ 0. Then
It should be easy enough to see that
• If n is odd, then n = 2k + 1 for some k ≥ 0. Then
so that
So, the overall series solution is
Z: is the number
[Z*(-1)]+(-50)
(-Z)+(-50)
-Z-50
<span>I hope it helps you :)</span>
Answer:
Oh well Is that it explain more
Step-by-step explanation:
Answer:
11
Step-by-step explanation:
it says 11