
MathGiven 1, 1/2, 1/6, 1/24, 1/120 In this sequence the numerator of each term is 1. The terms in the denominator follow the n! (factorial) pattern. The expression for the nth them of the sequence is...

MathThe `n`th term is `a_n=(1)^(n1)` Let us check: `a_1=(1)^0=1` `a_2=(1)^1=1` `a_3=(1)^2=1` `a_4=(1)^3=1` The solution is obvious if we know that `(1)^n=1` when `n` is even and `(1)^n=1`...

MathLet us start with the sequence `1,1,1,1,1,1,...` The `n`th term of this sequence is `(1)^n` (for odd `n` we get `1` while for even `n` we get `1`) If we add `2` to each term of this sequence...

MathLet us first write the first two terms as fractions as well. `1/1,3/1,3^2/2,3^3/6,3^4/24,3^5/120,...` In the numerator we have powers of 3 (`3^0=1` and `3^1=3`). In the denominator we have...

Math`T_1 = 1+1/2` `T_2 = 1+3/4` `T_3 = 1+7/8` `T_4 = 1+15/16` `T_5 = 1+31/32` Every term has a '1' in its expression. So the nth term will also have '1' as the beginning value. When we consider the...

Math`a_1=2^1/3^1=2/3` `a_2=2^2/3^2=4/9` `a_3=2^3/3^3=8/27` `a_4=2^4/3^4=16/81` `a_5=2^5/3^5=32/243`

Math`a_1=1/1^(3/2)=1/1=1` `a_2=1/2^(3/2)=1/sqrt(8)=1/(2sqrt2)` `a_3=1/3^(3/2)=1/sqrt27=1/(3sqrt3)` `a_4=1/4^(3/2)=1/sqrt64=1/(4sqrt4)` `a_5=1/5^(3/2)=1/sqrt125=1/(5sqrt5)`

MathThis is a constant sequence, meaning that all the terms are the same. Therefore, the first five terms are the same. In this case they are equal to `2/3.` `a_1=2/3` `a_2=2/3` `a_3=2/3` `a_4=2/3`...

Math`a_1=1+(1)^1=11=0` `a_2=1+(1)^2=1+1=2` `a_3=1+(1)^3=11=0` `a_4=1+(1)^4=1+1=2` `a_5=1+(1)^5=11=0` This sequence alternates between 0 and 2.

Math`a_1=1(11)(12)=1cdot0cdot(1)=0` `a_2=2(21)(22)=2cdot1cdot0=0` `a_3=3(31)(32)=3cdot2cdot1=6` `a_4=4(41)(42)=4cdot3cdot2=24` `a_5=5(51)(52)=5cdot4cdot3=60`

Math`a_1=1(1^26)=5` `a_2=2(2^26)=2(2)=4` `a_3=3(3^26)=3cdot3=9` `a_4=4(4^26)=4cdot10=40` `a_5=5(5^26)=5cdot19=95`

Math`a_n=(1)^n(n/(n+1))` `a_1=(1)^1(1/(1+1))=(1)(1/2)=1/2` Plug in n=2, to get the 2nd term `a_2=(1)^2(2/(2+1))=(1)(1)(2/3)=2/3` Plug in n=3, to get the 3rd term...

Math`a_n=(1)^(n+1)/(n^2+1)` `a_1=(1)^(1+1)/(1^2+1)=1/2` `a_2=(1)^(2+1)/(2^2+1)=1/5` `a_3=(1)^(3+1)/(3^2+1)=1/10` `a_4=(1)^(4+1)/(4^2+1)=1/17` `a_5=(1)^(5+1)/(5^2+1)=1/26` ` ` ` `

MathIt is obvious that each term is greater by four than the previous term which implies that this is arithmetic sequence. And since the first term is 3 we have `a_n=3+4(n1)=3+4n4=1+4n` The `n`th...

MathLet us compare this sequence with the following one `1,4,9,16,25,...` This is the sequence of square numbers `n^2` and each term of our sequence is one smaller then the terms in the above sequence....

MathNumerator starts with 2 and is increased by one in each term that follows `(n+1).`Denominator starts with 3 and is increased by one in each term that follows `(n+2).` Signs alternate between + and...

MathThe expression for the apparent term is `a_n=(1)^(n1)/(2^n)` ` ` `a_1=(1)^(11)/2^1=1/2` `a_2=(1)^(21)/2^2=1/4` `a_3=(1)^(31)/2^3=1/8` `a_4=(1)^(41)/2^4=1/16`

MathThe numerators are an arithmetic sequence with the difference 1, the denominators  of the difference 2. The resulting formula is a_n= (n+1)/(2n1).

MathNumerator contains powers of 2 starting with `2^0=1` while the denominator contains powers of 3 starting with `3^1=3.` Therefore, the `n`th term is `a_n=2^(n1)/3^n`...

MathNumerator is always 1 while the denominator is a square number, thus we can write the sequence as follows `1/1^2,1/2^2,1/3^2,1/4^2,1/5^2,...` From this we see that the `n`th terms is `a_n=1/n^2`

Math`a_n=4n7` `a_1=4*17=3` `a_2=4*27=87=1` `a_3=4*37=127=5` `a_4=4*47=167=9` `a_5=4*57=207=13` So, the first five terms of the sequence are `3,1,5,9,13`

Math`3^n` is 3, 9, 27, 81 and 243. Therefore the first five terms are 21/3 = 1 and 2/3, 21/9 = 1 and 8/9, 21/27 = 1 and 26/27, 21/81 = 1 and 80/81, 21/243 = 1 and 242/243.

Math`a_n=(2)^n` `a_1=(2)^1=2` `a_2=(2)^2=2*2=4` `a_3=(2)^3=2*2*2=8` `a_4=(2)^4=2*2*2*2=16` `a_5=(2)^5=2*2*2*2*2=32` So, the first five terms of the sequence are `2,4,8,16,32`

Math`a_n=(1/2)^n` `a_1=(1/2)^1=1/2` `a_2=(1/2)^2=1/4` `a_3=(1/2)^3=1/8` `a_4=(1/2)^4=1/16` `a_5=(1/2)^5=1/32` So, the first five terms of the sequence are `1/2,1/4,1/8,1/16,1/32`

Math`a_n=n/(n+2)` `a_1=1/(1+2)=1/3` `a_2=2/(2+2)=2/4=1/2` `a_3=3/(3+2)=3/5` `a_4=4/(4+2)=4/6=2/3` `a_5=5/(5+2)=5/7` So, the first five terms of the sequence are `1/3,1/2,3/5,2/3,5/7`

MathThese terms are fractions, their numerators and denominators are easy to compute. The first five terms are 3, 12/11, 9/13, 24/47 and 15/37.

Math`(1)^n` is 1 for even n and 1 for odd n. So `1+(1)^n` is 2 or 0, respectively. So the first five terms are 0, 1, 0, 1/2 and 0.

MathThe first five terms are 1, 1/4, 1/9, 1/16 and 1/25.

Math`2x+6y=16` `2x3y=7` Write the system of equations as an augmented matrix. `[[2,616],[2,37]]` Multiply the second line by 1. `[[2,616],[2,37]]` Eliminate the first column....

MathrsWe began by stating our equations as follows: 3x  2y = 27 x + 3y = 13 (NB: Note that I wrote the values of the x variable below each other, values with the y variable below each other and the...

Math`x+y=4` `2x4y=34` Write the system of equations as an augmented matrix. `[[1,14],[2,434]]` Multiply the first line by a 2. `[[2,28],[2,434]] ` Eliminate the first column....

Math`x+2y3z=28` `4y+2z=0` `x+yz=5` The equations in the matrix form can be written as, `[[1,2,3,28],[0,4,2,0],[1,1,1,5]]` Add Row 1 and Row 3 `[[1,2,3,28],[0,4,2,0],[0,3,4,33]]` Multiply...

Math`3x2y+z=15` `x+y+2z=10` `xy4z=14` `A=[[3,2,1],[1,1,2],[1,1,4]]` `b=[[15],[10],[14]]` `[Ab]=[[3,2,1,15],[1,1,2,10],[1,1,4,14]]` Multiply 2nd Row by 3 and add Row 1...

MathThe augmented matrix is `[[1,1,22],[3,4,4],[4,8,32]]` On applying `R_1 gt R_1 ` and `R_3 gt (R_3)/4` we get (means inverse the sign of first row and divide the third row by 4)...

MathThe augmented matrix `[[1,2,0],[1,1,6],[3,2,8]]` On applying `R_2 gt R_1  R_2 ` we get `[[1,2,0],[0,1,6],[3,2,8]]` On applying `R_3 gt 3R_1  R_3 ` we get `[[1,2,0],[0,1,6],[0,8,8]]` On...

Math`3x+2yz+w=0` `xy+4z+2w=25` `2x+y+2zw=2` `x+y+z+w=6` The above system of equations can be represented by the coefficient matrix A and right hand side vector b as follows:...

Math`x4y+3z2w=9` `3x2y+z4w=13` `4x+3y2z+w=4` `2x+y4z+3w=10` The above system of equations can be represented by the coefficient matrix A and right hand side matrix b as follows:...

MathGiven , 2x + 6y = 22, x + 2y = 9 so A = 2 6 1 2 and B = 22 9 so the augmented matrix is A^~=[A B]= 2 6 22 1 2 9 step 1....

MathGiven system of equations are 5x  5y = 5, 2x  3y = 7 so ,we get the matrices as A= 5 5 2 3 and B= 5 7 the augmented matrix [AB] = 5 5 5...

MathGiven system of equations are 8x  4y = 7, 5x + 2y = 1 so the matrix A ,B are given as foll A = 8 4 5 2 B= 7 1 so the augmented matrix is [AB] = 8 4 7...

MathGiven system of equations are , x  3y = 5, 2x + 6y = 10 so the matrix A,B are as follows, A = 1 3 2 6 and B = 5 10 so the augmented matrix is [AB] = 1 3 5...

MathThe augmented matrix is `[[1,2,1,8],[3,7,6,26]]` On applying `R_2 gt R_2  3R_1` we get `[[1,2,1,8],[0,1,3,2]]` On applying `R_1 gt R_1  2R_2 ` we get `[[1,0,5,4],[0,1,3,2]]` The corresponding...

MathThe augmented matrix is `[[1,1,4,5],[2,1,1,9]]` On applying `R_1 gt 2R_1  R_2` we get `[[1,1,4,5],[0,1,9,1]] ` On applying `R_1 gt R_1  R_2` we get `[[1,0,5,4],[0,1,9,1]]` The corresponding...

Math`x3z=2` `3x+y2z=5` `2x+2y+z=4` The equations can be written as `[[1,0,3,2],[3,1,2,5],[2,2,1,4]]` R2 `>` (R1+R3)R2 `[[1,0,3,2],[0,1,0,3],[2,2,1,4]]` R3 `>` (R32R2)...

Math`2xy+3z=24` `2yz=14` `7x5y=6` Write the equations as, `[[2,1,3,24],[0,2,1,14],[7,5,0,6]]` Make the pivot in the first column by dividing the First row by 2,...

MathThe augmented matrix is `[[1,1,1,14], [2,1,1,21],[3,2,1,19]] ` On applying `R_1 gt R_1 +R_2` we get (means changing 1st row as the sum of of first and second row)...

Math`2x+2yz=2` `x3y+z=28` `x+y=14` The equations can be written as, `[[2,2,1,2],[1,3,1, 28],[1,1,0,14]]` R2`>` R1+R2+R3 `[[2,2,1,2],[2,0,0,12],[1,1,0,14]]` R1 `>` R1R2...

MathThe order of a matrix is determined by the number of rows and the number of columns. This matrix has 2 rows and 2 columns. Therefore the order of the matrix is 2 X 2.

MathThe order of a matrix is determined by the number of rows and the number of columns. This matrix has 2 rows and 3 columns. Therefore the order of the matrix is 2 X 3.

MathThe order of a matrix is determined by the number of rows and the number of columns. This matrix has 3 rows and 3 columns. Therefore the order of the matrix is 3 X 3.
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