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sum_(i=2)^(4)5(3u_(i)+y_(i))

∑i=245(3ui+yi) \sum_{i=2}^{4} 5\left(3 u_{i}+y_{i}\right)

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Full solution

Q. ∑i=245(3ui+yi) \sum_{i=2}^{4} 5\left(3 u_{i}+y_{i}\right)
  1. Expand Series Term: First, let's expand the series term by term.\newlineFor i=2i=2: 5(3u2+y2)5(3u_2 + y_2)\newlineFor i=3i=3: 5(3u3+y3)5(3u_3 + y_3)\newlineFor i=4i=4: 5(3u4+y4)5(3u_4 + y_4)\newlineNow, we add them up.
  2. Addition of Terms: So, the sum is 5(3u2+y2)+5(3u3+y3)+5(3u4+y4)5(3u_2 + y_2) + 5(3u_3 + y_3) + 5(3u_4 + y_4). We can factor out the 55 to make it easier. 5[(3u2+y2)+(3u3+y3)+(3u4+y4)]5[(3u_2 + y_2) + (3u_3 + y_3) + (3u_4 + y_4)]
  3. Factor Out Common Factor: Now, we just need to plug in the values for uiu_i and yiy_i and calculate the sum. But wait, we don't have the actual values for uiu_i and yiy_i, so we can't calculate the final sum.

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