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sum_(k=0)^(19)-4(-2)^(k)~~ Choose 1 answer: (A) -699,052 (B) 1,398,100 (c) 2,097,152 (D) 4,194,308

∑k=019−4(−2)k≈ \sum_{k=0}^{19}-4(-2)^{k} \approx \newlineChoose 11 answer:\newline(A) −699,052 -699,052 \newline(B) 1,398,100 1,398,100 \newline(C) 2,097,152 2,097,152 \newline(D) 4,194,308 4,194,308

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Full solution

Q. ∑k=019−4(−2)k≈ \sum_{k=0}^{19}-4(-2)^{k} \approx \newlineChoose 11 answer:\newline(A) −699,052 -699,052 \newline(B) 1,398,100 1,398,100 \newline(C) 2,097,152 2,097,152 \newline(D) 4,194,308 4,194,308
  1. Identify series type: Identify the type of series.\newlineThe series is of the form −4(−2)k-4(-2)^k, which is a geometric series because each term is obtained by multiplying the previous term by a common ratio.
  2. Determine first term: Determine the first term of the series.\newlineThe first term is obtained when k=0k=0, which gives us −4(−2)0=−4(1)=−4-4(-2)^0 = -4(1) = -4.
  3. Determine common ratio: Determine the common ratio of the series.\newlineThe common ratio is −2-2, as each term is multiplied by −2-2 to get the next term.
  4. Use formula for sum: Use the formula for the sum of a finite geometric series.\newlineThe sum SS of a geometric series with first term aa, common ratio rr, and nn terms is given by S=a(1−rn)(1−r)S = \frac{a(1 - r^n)}{(1 - r)}, provided that r≠1r \neq 1.
  5. Apply formula to series: Apply the formula to the given series.\newlineHere, a=−4a = -4, r=−2r = -2, and n=20n = 20 (since we are summing from k=0k=0 to k=19k=19, which gives us 2020 terms).\newlineS=−4(1−(−2)20)/(1−(−2))S = -4(1 - (-2)^{20}) / (1 - (-2))
  6. Calculate sum: Calculate the sum.\newlineS=−4(1−220)/(1+2)S = -4(1 - 2^{20}) / (1 + 2)\newlineS=−4(1−1048576)/3S = -4(1 - 1048576) / 3\newlineS=−4(−1048575)/3S = -4(-1048575) / 3\newlineS=4×1048575/3S = 4 \times 1048575 / 3\newlineS=4194300/3S = 4194300 / 3
  7. Final calculation: Final calculation.\newlineS=1398100S = 1398100

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