S_n &= \displaystyle \sum_{k = 1}^n \left[a + (k - 1) d\right] r^{k-1} \\\\ S(1-r)& = a+dr&+dr^2&+\cdots+dr^{n-1}&-[a+(n-1)d]r^{n}.\\ \dfrac{1}{4} \left( 1+\dfrac{3}{2}+ \dfrac{5}{4}+\dfrac{7}{8}+ \cdots \right) .41(1+23+45+87+⋯). as a decimal notation, e.g. Let's practice with more examples: Calculate the value of the sum ∑i=1∞i7i\displaystyle\sum_{i=1}^{\infty} \dfrac{i}{7^i}i=1∑∞7ii. ∑i=1∞2i−12i+1=∑i=1∞(2i2i+1−12i+1)=∑i=1∞i2i−∑i=1∞12i+1.\sum_{i=1}^{\infty} \dfrac{2i-1}{2^{i+1}}=\displaystyle\sum_{i=1}^{\infty} \left(\dfrac{2i}{2^{i+1}}-\dfrac{1}{2^{i+1}}\right)=\displaystyle\sum_{i=1}^{\infty}\dfrac{i}{2^i}-\displaystyle\sum_{i=1}^{\infty}\dfrac{1}{2^{i+1}}.i=1∑∞2i+12i−1=i=1∑∞(2i+12i−2i+11)=i=1∑∞2ii−i=1∑∞2i+11. Using the formula for the sum of terms of an AGP for the expression in the parenthesis, we get. What is the expected number of coin flips before we get the first head? Fraction format button is used to work with all fractions. S=a+(a+d)r+(a+2d)r2+⋯+[a+(n−1)d]rn−1Sr=0+ar+(a+d)r2+⋯+[a+(n−2)d]rn−1+[a+(n−1)d]rnS(1−r)=a+dr+dr2+⋯+drn−1−[a+(n−1)d]rn.\begin{array} { rlllllllll} x[(n−1)xn−nxn−1+1]xn(1−x)2+(1x)[(n−1)(1x)n−n(1x)n−1+1](1x)n[1−(1x)]2+n.\dfrac{x [ (n-1)x^n-nx^{n-1}+1]}{x^n(1-x)^2} +\dfrac{\left( \frac 1x \right) \left[ (n-1)\left( \frac 1x \right)^n-n\left( \frac 1x \right)^{n-1}+1 \right]}{\left( \frac 1x \right)^n \left[ 1-\left( \frac 1x \right) \right]^2} +n.xn(1−x)2x[(n−1)xn−nxn−1+1]+(x1)n[1−(x1)]2(x1)[(n−1)(x1)n−n(x1)n−1+1]+n. Given that the infinite sum S SS can be expressed as ab \frac abba, where aaa and bbb are coprime positive integers, find a+ba+ba+b. 2S=1⋅22+2⋅23+⋯+99⋅2100+100⋅2101.2S= 1\cdot 2^2+2 \cdot 2^3+\cdots+99 \cdot 2^{100}+100 \cdot 2^{101}.2S=1⋅22+2⋅23+⋯+99⋅2100+100⋅2101. Let's see if you can solve the following problem. limn→∞rn=0. Taking the limit as nnn goes to infinity, we get, limn→∞Sn=a−01−r+dr(1−0)(1−r)2=a1−r+dr(1−r)2, where ∣r∣<1. S∞=1−ra+(1−r)2dr. -S&=2\dfrac{(2^{100}-1)}{2-1}-100 \cdot 2^{101} \qquad (\text{since the first 100 terms are in GP})\\ \frac{1}{2} T = \frac{1}{2} + \frac{ \frac{2}{4} } { 1 - \frac{1}{2} } = \frac{1}{2} + 1 = \frac{3}{2} . The terms of this sequence are too large for us to want to attempt to sum them manually. Calculate the value of the summation ∑i=1∞2i−12i+1\displaystyle\sum_{i=1}^{\infty} \dfrac{2i-1}{2^{i+1}}i=1∑∞2i+12i−1. their second finite difference is a constant. Note: the roll of the 6 is included in the "number of throws before we get the first 6.". 1xn(x+2x2+⋯+(n−1)xn−1)+[xn−1+2xn−2+⋯+(n−1)x]+n.\dfrac{1}{x^{n}} \left( x+2x^2+\cdots+(n-1)x^{n-1} \right) +\left[ x^{n-1}+2x^{n-2}+\cdots+(n-1)x \right] +n.xn1(x+2x2+⋯+(n−1)xn−1)+[xn−1+2xn−2+⋯+(n−1)x]+n. S&=100 \cdot 2^{101}-2 \cdot 2^{100}+2\\ S=12+24+38+416+532+⋯ .S=\dfrac 12 +\dfrac 24 +\dfrac 38+\dfrac{4}{16}+\dfrac{5}{32}+\cdots.S=21+42+83+164+325+⋯. S&=\dfrac 12 &+\dfrac 24 &+\dfrac 38 &+\dfrac{4}{16} &+\dfrac{5}{32}+ \cdots \\ \ _\square We will use a different approach to reduce this to a "linear-geometric progression", which is an AGP. Now let's derive a general formula for the sum of terms of an AGP with initial term aaa, common difference ddd and common ratio rrr: The sum of the first nnn terms of an AGP is given by, Sn=∑k=1n[a+(k−1)d]rk−1=a−[a+(n−1)d]rn1−r+dr(1−rn−1)(1−r)2.\begin{aligned} Already have an account? \hline We thus get. Supercharge your algebraic intuition and problem solving skills! The calculator will provide the rest. □\displaystyle\sum_{n=1}^{\infty} nP(n)=\displaystyle\sum_{n=1}^{\infty} \dfrac{n}{2^n}=2. Let's find a more general approach, and we start by looking at an example. If we had to describe this summation, we would call it a "quadratic-geometric progression", because the numerator is a quadratic i2 i^2 i2. □S=2 \times \left( \dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}} \right)=2. \hline We then get 23U−1227=∑i=1∞63i+3, \frac{2}{3} U - \frac{12}{27} = \displaystyle \sum_{i=1}^\infty \frac{6}{3^{i+3} } ,32U−2712=i=1∑∞3i+36, which is a geometric progression with an infinite sum of 6811−13=19 \frac{ \frac{6}{81} } { 1 - \frac{1}{3} } = \frac{1}{9} 1−31816=91. \hline Arithmetic and Geometric Progressions: Problem Solving, https://brilliant.org/wiki/arithmetic-geometric-progression/. \end{aligned}Sn=k=1∑n[a+(k−1)d]rk−1=1−ra−[a+(n−1)d]rn+(1−r)2dr(1−rn−1).. Clearly, this is an infinite AGP with a=1,r=17,d=1a=1, r=\frac{1}{7}, d=1a=1,r=71,d=1. \end{array}S2SS(1−2)=1⋅2=0=1⋅2+2⋅22+1⋅22+1⋅22+3⋅23+2⋅23+1⋅23+⋯+100⋅2100+⋯+99⋅2100+⋯+1⋅2100+−100⋅2101100⋅2101., −S=2(2100−1)2−1−100⋅2101(since the first 100 terms are in GP)S=100⋅2101−2⋅2100+2S=200⋅2100−2⋅2100+2S=198⋅2100+2. New user? \frac{1}{2} T & = \frac{1}{2} & + \frac{2}{4} & + \frac{2}{8} & + \frac{2}{16} & + \frac{2}{32} & + \cdots .\\ i.e. In variables, it looks like. If the fraction or mixed number is only part of the calculation then omit clicking equals and continue with the calculation per usual. S&=200 \cdot 2^{100} - 2\cdot 2^{100}+2\\ S(1-2)& = 1 \cdot 2&+1 \cdot 2^2&+1 \cdot 2^3&+\cdots+1 \cdot 2^{100}&-&100 \cdot 2^{101}.\\ Hence, the total is 2−12=1.5 □ 2 - \frac{1}{2} = 1.5 \ _\square 2−21=1.5 □. We will be using the fact that in this case. It is assumed that all bonds pay interest semi-annually. S2=14+28+316+432+564+⋯ . \frac{1}{3} T & = & + \frac{7}{27} & + \frac{ 19}{81} & + \frac{ 37}{ 243} & + \cdots \\ S_\infty = \dfrac{ a } { 1 - r } + \dfrac{ dr } { (1-r)^2 }. S=(17)(1+27+372+473+⋯ ).S=\left( \dfrac 17 \right) \left( 1+\dfrac{2}{7}+\dfrac{3}{7^2}+\dfrac{4}{7^3}+\cdots \right) .S=(71)(1+72+723+734+⋯). Clarification: 2,5,9,14,20,…2,5,9,14,20,\ldots2,5,9,14,20,… follows a 2nd2^{\text{nd}}2nd-degree polynomial function, i.e. \end{array} S31S32S=31==31+98+91+97+2727+278+2719+8164+8127+8137+243125+24364+24361+⋯+⋯+⋯., We thus get 23S−13=∑i=1∞3i2+3i+13i+1 \frac{2}{3} S - \frac{1}{3} =\displaystyle \sum_{i=1}^\infty \frac{ 3i^2+3i+1}{3^{i+1} } 32S−31=i=1∑∞3i+13i2+3i+1, which is a "quadratic-geometric progression". To enhance your problem solving skills on arithmetic progressions, geometric progressions, and arithmetic-geometric progression, you can check out the following wikis: Learn more in our Algebra Fundamentals course, built by experts for you. □. □\begin{aligned} Let's try one problem to practice the above method: 1+2⋅2+3⋅22+4⋅23+⋯+100⋅299= ?1+2 \cdot 2+ 3 \cdot 2^2 + 4 \cdot 2^3 + \cdots+ 100 \cdot 2^{99}= \, ?1+2⋅2+3⋅22+4⋅23+⋯+100⋅299=? \frac{1}{2} T & = & + \frac 14 & +\frac {3}{8} & +\frac{5}{16} &+\frac{7}{32} &+\cdots \\ This holds more generally: When we take the difference of terms in a degree nnn sequence, we will get a degree n−1n-1 n−1 sequence. □. Writing down the given summation as the difference of two summations, we get. All rights reserved. In the following series, the numerators are in AP and the denominators are in GP: 12+24+38+416+532+⋯= ?\large \dfrac{\color{#3D99F6}{1}}{\color{#D61F06}{2}}+\dfrac{\color{#3D99F6}{2}}{\color{#D61F06}{4}}+\dfrac{\color{#3D99F6}{3}}{\color{#D61F06}{8}}+\dfrac{\color{#3D99F6}{4}}{\color{#D61F06}{16}}+\dfrac{\color{#3D99F6}{5}}{\color{#D61F06}{32}}+\cdots= \, ?21+42+83+164+325+⋯=?
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