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Exercise 5.2 · Q4

Q.Evaluate ∑k=110(3+2k)\displaystyle\sum_{k=1}^{10} (3+2^k).

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Split the sum into a constant sum and a geometric-series sum, then add.

For a G.P. with first term aa and common ratio r≠1r\neq1, sum of nn terms: Sn=a(rn−1)r−1S_n=\dfrac{a(r^n-1)}{r-1}.

  1. Split the sum: ∑k=110(3+2k)=∑k=1103+∑k=1102k\displaystyle\sum_{k=1}^{10}(3+2^k)=\sum_{k=1}^{10}3+\sum_{k=1}^{10}2^k.
  2. The first part is a constant added 10 times: ∑k=1103=3×10=30\sum_{k=1}^{10}3=3\times10=30. …

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