IB AI · Unit 2 · 2B Geometric Sequences and Series
Pick the sum formula from the information you already have. A summation in symbol form can go straight into the calculator.
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SL 1.3
\(u_n = u_1 r^{n-1}\)
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1},\ r \neq 1\)
\(S_n = \dfrac{u_1(1 - r^n)}{1 - r},\ r \neq 1\)
The sum of n terms
If you already know the first term and the last term, use
\(S_n = \dfrac{u_1 - r u_n}{1 - r}\), \(r \neq 1\)
If you know the first term and the common ratio, use
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\), \(r \neq 1\)
When \(r\) is less than 1, the other booklet form keeps the denominator positive.
\(S_n = \dfrac{u_1(1 - r^n)}{1 - r}\)
The first term and the common ratio are given.
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(u_1 = 3\)
\(r = 2\)
\(n = 5\)
\(S_{5} = \dfrac{3\left(2^{5} - 1\right)}{2 - 1}\)
\(2^{5} = 32\)
\(32 - 1 = 31\)
\(2 - 1 = 1\)
\(S_{5} = \dfrac{3 \times 31}{1}\)
\(S_{5} = 93\)
Answer: the sum = \(93\)
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(u_1 = 4\)
\(r = 3\)
\(n = 4\)
\(S_{4} = \dfrac{4\left(3^{4} - 1\right)}{3 - 1}\)
\(3^{4} = 81\)
\(81 - 1 = 80\)
\(3 - 1 = 2\)
\(S_{4} = \dfrac{4 \times 80}{2}\)
\(4 \times 80 = 320\)
\(S_{4} = \dfrac{320}{2}\)
\(S_{4} = 320 \div 2\)
\(S_{4} = 160\)
Answer: the sum = \(160\)
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(u_1 = 5\)
\(r = 2\)
\(n = 6\)
\(S_{6} = \dfrac{5\left(2^{6} - 1\right)}{2 - 1}\)
\(2^{6} = 64\)
\(64 - 1 = 63\)
\(2 - 1 = 1\)
\(S_{6} = \dfrac{5 \times 63}{1}\)
\(S_{6} = 315\)
Answer: the sum = \(315\)
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(u_1 = 6\)
\(r = 2\)
\(n = 5\)
\(S_{5} = \dfrac{6\left(2^{5} - 1\right)}{2 - 1}\)
\(2^{5} = 32\)
\(32 - 1 = 31\)
\(2 - 1 = 1\)
\(S_{5} = \dfrac{6 \times 31}{1}\)
\(S_{5} = 186\)
Answer: the sum = \(186\)
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(u_1 = 2\)
\(r = 3\)
\(n = 5\)
\(S_{5} = \dfrac{2\left(3^{5} - 1\right)}{3 - 1}\)
\(3^{5} = 243\)
\(243 - 1 = 242\)
\(3 - 1 = 2\)
\(S_{5} = \dfrac{2 \times 242}{2}\)
\(2 \times 242 = 484\)
\(S_{5} = \dfrac{484}{2}\)
\(S_{5} = 484 \div 2\)
\(S_{5} = 242\)
Answer: the sum = \(242\)
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(u_1 = 8\)
\(r = 2\)
\(n = 7\)
\(S_{7} = \dfrac{8\left(2^{7} - 1\right)}{2 - 1}\)
\(2^{7} = 128\)
\(128 - 1 = 127\)
\(2 - 1 = 1\)
\(S_{7} = \dfrac{8 \times 127}{1}\)
\(S_{7} = 1016\)
Answer: the sum = \(1016\)
The first term, the last term, and the common ratio are given.
\(S_n = \dfrac{u_1 - r u_n}{1 - r}\)
\(u_1 = 3\)
\(u_n = 48\)
\(r = 2\)
\(n = 5\)
\(S_{5} = \dfrac{3 - 2 \times 48}{1 - 2}\)
\(2 \times 48 = 96\)
\(S_{5} = \dfrac{-93}{-1}\)
\(S_{5} = 93\)
Answer: the sum = \(93\)
\(S_n = \dfrac{u_1 - r u_n}{1 - r}\)
\(u_1 = 4\)
\(u_n = 108\)
\(r = 3\)
\(n = 4\)
\(S_{4} = \dfrac{4 - 3 \times 108}{1 - 3}\)
\(3 \times 108 = 324\)
\(S_{4} = \dfrac{-320}{-2}\)
\(S_{4} = 160\)
Answer: the sum = \(160\)
\(S_n = \dfrac{u_1 - r u_n}{1 - r}\)
\(u_1 = 5\)
\(u_n = 160\)
\(r = 2\)
\(n = 6\)
\(S_{6} = \dfrac{5 - 2 \times 160}{1 - 2}\)
\(2 \times 160 = 320\)
\(S_{6} = \dfrac{-315}{-1}\)
\(S_{6} = 315\)
Answer: the sum = \(315\)
\(S_n = \dfrac{u_1 - r u_n}{1 - r}\)
\(u_1 = 6\)
\(u_n = 96\)
\(r = 2\)
\(n = 5\)
\(S_{5} = \dfrac{6 - 2 \times 96}{1 - 2}\)
\(2 \times 96 = 192\)
\(S_{5} = \dfrac{-186}{-1}\)
\(S_{5} = 186\)
Answer: the sum = \(186\)
\(S_n = \dfrac{u_1 - r u_n}{1 - r}\)
\(u_1 = 2\)
\(u_n = 162\)
\(r = 3\)
\(n = 5\)
\(S_{5} = \dfrac{2 - 3 \times 162}{1 - 3}\)
\(3 \times 162 = 486\)
\(S_{5} = \dfrac{-484}{-2}\)
\(S_{5} = 242\)
Answer: the sum = \(242\)
\(S_n = \dfrac{u_1 - r u_n}{1 - r}\)
\(u_1 = 8\)
\(u_n = 512\)
\(r = 2\)
\(n = 7\)
\(S_{7} = \dfrac{8 - 2 \times 512}{1 - 2}\)
\(2 \times 512 = 1024\)
\(S_{7} = \dfrac{-1016}{-1}\)
\(S_{7} = 1016\)
Answer: the sum = \(1016\)
Copy the summation into a TI-84 Plus CE.
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(3\times 2^{x-1}\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 5.
7. Press ENTER.
The calculator shows \(93\).
Answer: the sum = \(93\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(4\times 3^{x-1}\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 4.
7. Press ENTER.
The calculator shows \(160\).
Answer: the sum = \(160\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(5\times 2^{x-1}\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 6.
7. Press ENTER.
The calculator shows \(315\).
Answer: the sum = \(315\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(2\times 3^{x-1}\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 4.
7. Press ENTER.
The calculator shows \(80\).
Answer: the sum = \(80\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(6\times 2^{x-1}\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 5.
7. Press ENTER.
The calculator shows \(186\).
Answer: the sum = \(186\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(8\times 2^{x-1}\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 4.
7. Press ENTER.
The calculator shows \(120\).
Answer: the sum = \(120\)
Find the missing first term or common ratio, then the sum.
\(r = \dfrac{u_2}{u_1}\)
\(r = \dfrac{6}{3}\)
\(r = 2\)
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(S_{5} = \dfrac{3\left(2^{5} - 1\right)}{2 - 1}\)
\(2^{5} = 32\)
\(32 - 1 = 31\)
\(2 - 1 = 1\)
\(S_{5} = \dfrac{3 \times 31}{1}\)
\(S_{5} = 93\)
Answer: the sum = \(93\)
\(r = \dfrac{u_2}{u_1}\)
\(r = \dfrac{12}{4}\)
\(r = 3\)
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(S_{4} = \dfrac{4\left(3^{4} - 1\right)}{3 - 1}\)
\(3^{4} = 81\)
\(81 - 1 = 80\)
\(3 - 1 = 2\)
\(S_{4} = \dfrac{4 \times 80}{2}\)
\(4 \times 80 = 320\)
\(S_{4} = \dfrac{320}{2}\)
\(S_{4} = 320 \div 2\)
\(S_{4} = 160\)
Answer: the sum = \(160\)
\(r = \dfrac{u_2}{u_1}\)
\(r = \dfrac{10}{5}\)
\(r = 2\)
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(S_{6} = \dfrac{5\left(2^{6} - 1\right)}{2 - 1}\)
\(2^{6} = 64\)
\(64 - 1 = 63\)
\(2 - 1 = 1\)
\(S_{6} = \dfrac{5 \times 63}{1}\)
\(S_{6} = 315\)
Answer: the sum = \(315\)
\(r = \dfrac{u_2}{u_1}\)
\(r = \dfrac{6}{2}\)
\(r = 3\)
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(S_{5} = \dfrac{2\left(3^{5} - 1\right)}{3 - 1}\)
\(3^{5} = 243\)
\(243 - 1 = 242\)
\(3 - 1 = 2\)
\(S_{5} = \dfrac{2 \times 242}{2}\)
\(2 \times 242 = 484\)
\(S_{5} = \dfrac{484}{2}\)
\(S_{5} = 484 \div 2\)
\(S_{5} = 242\)
Answer: the sum = \(242\)
The second term is the first term multiplied by the common ratio.
\(u_2 = u_1 r\)
\(12 = u_1 \times 3\)
\(u_1 = 4\)
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(S_{4} = \dfrac{4\left(3^{4} - 1\right)}{3 - 1}\)
\(3^{4} = 81\)
\(81 - 1 = 80\)
\(3 - 1 = 2\)
\(S_{4} = \dfrac{4 \times 80}{2}\)
\(4 \times 80 = 320\)
\(S_{4} = \dfrac{320}{2}\)
\(S_{4} = 320 \div 2\)
\(S_{4} = 160\)
Answer: the sum = \(160\)
The second term is the first term multiplied by the common ratio.
\(u_2 = u_1 r\)
\(10 = u_1 \times 2\)
\(u_1 = 5\)
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(S_{6} = \dfrac{5\left(2^{6} - 1\right)}{2 - 1}\)
\(2^{6} = 64\)
\(64 - 1 = 63\)
\(2 - 1 = 1\)
\(S_{6} = \dfrac{5 \times 63}{1}\)
\(S_{6} = 315\)
Answer: the sum = \(315\)
The second term is the first term multiplied by the common ratio.
\(u_2 = u_1 r\)
\(16 = u_1 \times 2\)
\(u_1 = 8\)
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
\(S_{5} = \dfrac{8\left(2^{5} - 1\right)}{2 - 1}\)
\(2^{5} = 32\)
\(32 - 1 = 31\)
\(2 - 1 = 1\)
\(S_{5} = \dfrac{8 \times 31}{1}\)
\(S_{5} = 248\)
Answer: the sum = \(248\)
If you can write a formula for uₙ, the summation key adds the first n terms.
\(u_n = 3 \times 2^{n}\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(3\times 2^x\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 5.
7. Press ENTER.
The calculator shows \(186\).
Answer: the sum = \(186\)
\(u_n = 4 \times 3^{n-1}\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(4\times 3^{(x-1)}\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 4.
7. Press ENTER.
The calculator shows \(160\).
Answer: the sum = \(160\)
\(u_n = 5 \times 2^{n-1}\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(5\times 2^{(x-1)}\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 6.
7. Press ENTER.
The calculator shows \(315\).
Answer: the sum = \(315\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 6\)
\(r = 2\)
\(u_n = 6 \times 2^{n-1}\)
\(u_n = 3 \times 2 \times 2^{n-1}\)
\(u_n = 3 \times 2^n\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(3\times 2^x\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 5.
7. Press ENTER.
The calculator shows \(186\).
Answer: the formula is \(u_n = 3 \times 2^n\) and the sum = \(186\)
\(r = \dfrac{u_2}{u_1}\)
\(r = \dfrac{12}{4}\)
\(r = 3\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 4\)
\(r = 3\)
\(u_n = 4 \times 3^{n-1}\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(4\times 3^{(x-1)}\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 4.
7. Press ENTER.
The calculator shows \(160\).
Answer: the formula is \(u_n = 4 \times 3^{n-1}\) and the sum = \(160\)
\(u_1 = 10 \div 2\)
\(u_1 = 5\)
\(u_n = u_1 r^{n-1}\)
\(u_1 = 5\)
\(r = 2\)
\(u_n = 5 \times 2^{n-1}\)
On a TI-84 Plus CE:
1. Press MATH.
2. Press 0 to open summation Σ(.
3. Type the expression \(5\times 2^{(x-1)}\).
4. Move to the variable box and type \(x\).
5. Move to the lower limit and type 1.
6. Move to the upper limit and type 6.
7. Press ENTER.
The calculator shows \(315\).
Answer: the formula is \(u_n = 5 \times 2^{n-1}\) and the sum = \(315\)
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