Q: What is the sum of the 17th row of pascals triangle?

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depends. If you start Pascals triangle with (1) or (1,1). The fifth row with then either be (1,4,6,4,1) or (1,5,10,10,5,1). The sums of which are respectively 16 and 32.

Each number in Pascal's triangle is used twice when calculating the row below. Consequently the row total doubles with each successive row. If the row containing a single '1' is row zero, then T = 2r where T is the sum of the numbers in row r. So for r=100 T = 2100 = 1267650600228229401496703205376

The sum of the numbers in each row of Pascal's triangle is twice the sum of the previous row. Perhaps you can work it out from there. (Basically, you should use powers of 2.)

AnswerPascal's triangle is a triangular array of the binomial coefficients in a triangle. It is named after the French mathematician Blaise Pascal. It is mainly used in probability and algebra.1 Row 01 1 Row 11 2 1 Row 21 3 3 1 Row 31 4 6 4 1 Row 4 etc.Each number in the triangle is the sum of the two directly above it. The value of a row, if each entry is considered a decimal place, is a power of 11. So, in row 2, '1,2,1' becomes 112, and '1,5,10,10,5,1' (which will be in row 5) becomes, after carrying , 161,051 which is 115.

the horizontal sums doubles each time the sum of row 1 = 1 row 2= 2 row 3 = 4 row 4 = 8 row 5 = 16 etc etc.......... the horizontal sums doubles each time the sum of row 1 = 1 row 2= 2 row 3 = 4 row 4 = 8 row 5 = 16 etc etc..........

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The sum is 24 = 16

depends. If you start Pascals triangle with (1) or (1,1). The fifth row with then either be (1,4,6,4,1) or (1,5,10,10,5,1). The sums of which are respectively 16 and 32.

Sum of numbers in a nth row can be determined using the formula 2^n. For the 100th row, the sum of numbers is found to be 2^100=1.2676506x10^30.

The Fifth row of Pascal's triangle has 1,4,6,4,1. The sum is 16. Formula 2n-1 where n=5 Therefore 2n-1=25-1= 24 = 16.

Each number in Pascal's triangle is used twice when calculating the row below. Consequently the row total doubles with each successive row. If the row containing a single '1' is row zero, then T = 2r where T is the sum of the numbers in row r. So for r=100 T = 2100 = 1267650600228229401496703205376

The Fifth row of Pascal's triangle has 1,4,6,4,1. The sum is 16.

The sum of the numbers on the fifteenth row of Pascal's triangle is 215 = 32768.

The sum of the numbers in each row of Pascal's triangle is twice the sum of the previous row. Perhaps you can work it out from there. (Basically, you should use powers of 2.)

Each element of a row of pascal's triangle is the sum of the two elements above it. Therefore when you some the elements of a row, each of the elements of the row above are being summed twice. Thus the sum of each row of pascal's triangle is twice the sum of the previous row.

AnswerPascal's triangle is a triangular array of the binomial coefficients in a triangle. It is named after the French mathematician Blaise Pascal. It is mainly used in probability and algebra.1 Row 01 1 Row 11 2 1 Row 21 3 3 1 Row 31 4 6 4 1 Row 4 etc.Each number in the triangle is the sum of the two directly above it. The value of a row, if each entry is considered a decimal place, is a power of 11. So, in row 2, '1,2,1' becomes 112, and '1,5,10,10,5,1' (which will be in row 5) becomes, after carrying , 161,051 which is 115.

Let's calculate the sums of few first rows of Pascal's triangle: 1st row: 1 = 1 2nd row: 1 + 1 = 2 3rd row: 1 + 2 + 1 = 4 Looks promising, let's continue: 4th row: 1 + 3 + 3 + 1 = 8 5th row: 1 + 4 + 6 + 4 + 1 = 16 We can make an assumption that each row's sum is twice the sum of previous row - it's a power of two. But why is that? If you know how Pascal's triangle is constructed, you should notice that when creating new row, you use the previous row numbers(except ones) two times in addition. Considering ones, you only use each 1 in previous row once, but in the new row you always add two 1's on the sides. Alternatively, you may think of empty space around Pascal's triangle as zeros and then you'll definitely use each previous row's numbers two times to create a new row. The formula will be then: s = 2n-1, where s - sum of the nth row(we assume numeration starts with 1 for the single '1' on the top) n - number of the row

the horizontal sums doubles each time the sum of row 1 = 1 row 2= 2 row 3 = 4 row 4 = 8 row 5 = 16 etc etc.......... the horizontal sums doubles each time the sum of row 1 = 1 row 2= 2 row 3 = 4 row 4 = 8 row 5 = 16 etc etc..........