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Evaluate the summation below.

7sum_(k=0)^(4)(4k-2k^(2))
Answer:

Evaluate the summation below.\newline7k=04(4k2k2) 7 \sum_{k=0}^{4}\left(4 k-2 k^{2}\right) \newlineAnswer:

Full solution

Q. Evaluate the summation below.\newline7k=04(4k2k2) 7 \sum_{k=0}^{4}\left(4 k-2 k^{2}\right) \newlineAnswer:
  1. Evaluate Series: We need to evaluate the summation of the series given by the expression 77 times the sum from k=0k = 0 to 44 of (4k2k2)(4k - 2k^2). We will do this by calculating the sum term by term and then multiplying by 77.
  2. Calculate Terms: First, let's calculate the sum without the factor of 77. We will substitute k=0,1,2,3,k = 0, 1, 2, 3, and 44 into the expression (4k2k2)(4k - 2k^2) and find the sum of these values.
  3. Calculate Sum: For k=0k = 0: (4×02×02)=0(4\times0 - 2\times0^2) = 0. For k=1k = 1: (4×12×12)=42=2(4\times1 - 2\times1^2) = 4 - 2 = 2. For k=2k = 2: (4×22×22)=88=0(4\times2 - 2\times2^2) = 8 - 8 = 0. For k=3k = 3: (4×32×32)=1218=6(4\times3 - 2\times3^2) = 12 - 18 = -6. For k=4k = 4: (4×42×42)=1632=16(4\times4 - 2\times4^2) = 16 - 32 = -16. Now, we add these values together to get the sum.
  4. Multiply by 77: The sum is 0+2+0616=200 + 2 + 0 - 6 - 16 = -20.
  5. Multiply by 77: The sum is 0+2+0616=200 + 2 + 0 - 6 - 16 = -20.Now, we multiply this sum by 77 to get the final answer.\newline7×(20)=1407 \times (-20) = -140.

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