A-Level Edexcel Maths Pure Modelling involving Numerical Methods: (i) Show that \[ \sum

(i) Show that \[ \sum_{r=1}^{16} (3 + 5r + 2^r) = 131798 \] (ii) A sequence $ u_1, u_2, u_3, \dots $ is defined by \[ u_{n+1} = \frac{1}{u_n}, \quad u_1 = \frac{2}{3} \] Find the exact value of \[ \sum_{r=1}^{100} u_r \] - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 2

Question 5

$(i)-Show-that--\[-\sum_{r=1}^{16}-(3-+-5r-+-2^r)-=-131798-\]--(ii)-A-sequence-$-u_1,-u_2,-u_3,-\dots-$-is-defined-by--\[-u_{n+1}-=-\frac{1}{u_n},-\quad-u_1-=-\frac{2}{3}-\]--Find-the-exact-value-of--\[-\sum_{r=1}^{100}-u_r-\]-Edexcel-A-Level Maths Pure-Question 5-2018-Paper 2.png$

(i) Show that \[ \sum_{r=1}^{16} (3 + 5r + 2^r) = 131798 \] (ii) A sequence $ u_1, u_2, u_3, \dots $ is defined by \[ u_{n+1} = \frac{1}{u_n}, \quad u_1 = \frac... show full transcript

Worked Solution & Example Answer:(i) Show that \[ \sum_{r=1}^{16} (3 + 5r + 2^r) = 131798 \] (ii) A sequence $ u_1, u_2, u_3, \dots $ is defined by \[ u_{n+1} = \frac{1}{u_n}, \quad u_1 = \frac{2}{3} \] Find the exact value of \[ \sum_{r=1}^{100} u_r \] - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 2

Step 1

Show that $ \sum_{r=1}^{16} (3 + 5r + 2^r) = 131798 $

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Answer

To find the sum ( \sum_{r=1}^{16} (3 + 5r + 2^r) ), we can split it into three separate sums:

Sum of constant 3: [ \sum_{r=1}^{16} 3 = 3 \times 16 = 48 ]
Sum of the linear term (5r): [ \sum_{r=1}^{16} 5r = 5 \sum_{r=1}^{16} r = 5 \times \frac{16(16+1)}{2} = 5 \times 136 = 680 ]
Sum of the exponential term (2^r): Recognizing this as a geometric series, [ \sum_{r=1}^{16} 2^r = 2(1 - 2^{16})/(1 - 2) = 2(2^{16} - 1) = 2^{17} - 2 = 131072 - 2 = 131070 ]

Combining all sums: [ \sum_{r=1}^{16} (3 + 5r + 2^r) = 48 + 680 + 131070 = 131798 ]

Step 2

Find the exact value of $ \sum_{r=1}^{100} u_r $

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Answer

Given that ( u_1 = \frac{2}{3} ) and ( u_{n+1} = \frac{1}{u_n} ), we can find further terms in the sequence:

( u_1 = \frac{2}{3} )
( u_2 = \frac{1}{u_1} = \frac{3}{2} )
( u_3 = \frac{1}{u_2} = \frac{2}{3} )

Notice that the sequence alternates: ( u_1 = \frac{2}{3}, u_2 = \frac{3}{2}, u_3 = \frac{2}{3}, u_4 = \frac{3}{2}, \dots )

Thus, we observe: [ u_{2n-1} = \frac{2}{3} \quad \text{and} \quad u_{2n} = \frac{3}{2} ]

To find ( \sum_{r=1}^{100} u_r ), we can calculate the sums of the odd and even indexed terms:

Odd indexed terms (50 times): [ 50 \times \frac{2}{3} = \frac{100}{3} ]
Even indexed terms (50 times): [ 50 \times \frac{3}{2} = 75 ]

Combining: [ \sum_{r=1}^{100} u_r = \frac{100}{3} + 75 = \frac{100}{3} + \frac{225}{3} = \frac{325}{3} ]

(i) Show that \[ \sum_{r=1}^{16} (3 + 5r + 2^r) = 131798 \] (ii) A sequence \( u_1, u_2, u_3, \dots \) is defined by \[ u_{n+1} = \frac{1}{u_n}, \quad u_1 = \frac{2}{3} \] Find the exact value of \[ \sum_{r=1}^{100} u_r \] - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 2

Show that \( \sum_{r=1}^{16} (3 + 5r + 2^r) = 131798 \)

Find the exact value of \( \sum_{r=1}^{100} u_r \)

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