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Evaluate the summation below. 3sum_(n=0)^(3)(-4n^(2)+3) Answer:

Evaluate the summation below.\newline3∑n=03(−4n2+3) 3 \sum_{n=0}^{3}\left(-4 n^{2}+3\right) \newlineAnswer:

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Q. Evaluate the summation below.\newline3∑n=03(−4n2+3) 3 \sum_{n=0}^{3}\left(-4 n^{2}+3\right) \newlineAnswer:
  1. Write Series for Each nn: Write down the series explicitly for each value of nn from 00 to 33.
    For n=0n = 0: 3(−4(0)2+3)=3(0+3)=93(-4(0)^2 + 3) = 3(0 + 3) = 9
    For n=1n = 1: 3(−4(1)2+3)=3(−4+3)=−33(-4(1)^2 + 3) = 3(-4 + 3) = -3
    For n=2n = 2: 3(−4(2)2+3)=3(−16+3)=−393(-4(2)^2 + 3) = 3(-16 + 3) = -39
    For nn00: nn11
  2. Calculate Series Results: Add the results of the series for each value of nn.
    Sum = 9+(−3)+(−39)+(−99)9 + (-3) + (-39) + (-99)
    Sum = 9−3−39−999 - 3 - 39 - 99
    Sum = 6−39−996 - 39 - 99
    Sum = −33−99-33 - 99
    Sum = −132-132

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