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A sequence is defined by $u_1 = a$ and $u_{n+1} = -1 \times u_n$ Find $$\sum_{n=1}^{95} u_n$$ Circle your answer. - AQA - A-Level Maths Pure - Question 3 - 2021 - Paper 2

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A-sequence-is-defined-by--$u_1-=-a$-and-$u_{n+1}-=--1-\times-u_n$-Find--$$\sum_{n=1}^{95}-u_n$$--Circle-your-answer.-AQA-A-Level Maths Pure-Question 3-2021-Paper 2.png

A sequence is defined by $u_1 = a$ and $u_{n+1} = -1 \times u_n$ Find $$\sum_{n=1}^{95} u_n$$ Circle your answer.

Worked Solution & Example Answer:A sequence is defined by $u_1 = a$ and $u_{n+1} = -1 \times u_n$ Find $$\sum_{n=1}^{95} u_n$$ Circle your answer. - AQA - A-Level Maths Pure - Question 3 - 2021 - Paper 2

Step 1

Find $u_n$ for $n = 1, 2, 3, ...$

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Answer

Given u1=au_1 = a, we can find the subsequent terms in the sequence using the recursive relation.

  • For n=1n=1: u2=−1×u1=−au_2 = -1 \times u_1 = -a
  • For n=2n=2: u3=−1×u2=−(−a)=au_3 = -1 \times u_2 = -(-a) = a
  • For n=3n=3: u4=−1×u3=−au_4 = -1 \times u_3 = -a

From this pattern, we can see that the terms alternate between aa and −a-a.

Step 2

Sum the first 95 terms

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Answer

The sequence for the first 95 terms will be:

  • u1=au_1 = a
  • u2=−au_2 = -a
  • u3=au_3 = a
  • u4=−au_4 = -a ... up to u95u_{95}.

This sequence consists of 48 terms of aa and 47 terms of −a-a because 95 is odd. Therefore, we can compute the sum:

∑n=195un=48a+47(−a)=48a−47a=a\sum_{n=1}^{95} u_n = 48a + 47(-a) = 48a - 47a = a

Step 3

Circle your answer

96%

101 rated

Answer

The final answer to circle is aa.

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