n=1 a(1) = 0.001 n=2 a(2) = 0.012 n=3 a(3) = 0.144 n=4 a(4) = 1.728 n=5 a(5) = 20.736 .... n=k a(k) = (12^(k-1))/1000 let n = k+1 a(k+1) = 12^(k)/1000 12(ak) = a(k+1) 12(12^k-1)/1000 = 12^k/1000 the 12 gets absorbed here. 12^(k-1+1)/1000 = 12^k/1000 Valid for k and k+1 therefore our equation A(n) = 12^(n-1)/1000, n(greater than or equal to) 1
To get a list of the squares of the first 1000 numbers we can do:> [n^2 | n sum [n^2 | n
There is an equation to find the sum of a list of numbers . It is called the Arithmetic Progression. The equation is S(n) = (n/2)[a + (n-1)d] 4 Where n = number of terms (1000) a = first term (1000) d = difference between terms (1) Hence S(1000) = (1000/2)[1000 + ( 1000 - 1)(1)] S(1000) = 500[ 1000 + 999] S(1000) = 500[1999] S(1000) = 999500 The answer!!!!!
1 N is equal to 1/1000 kN. 23400 is equal to 23400/1000=23.4 kN.
The formula for the nth triangular number is Tn = n(n+1)/2. To find the 1000th triangular number, we substitute n = 1000 into the formula: T1000 = 1000(1000+1)/2 = 1000(1001)/2 = 500500. Therefore, the 1000th triangular number is 500500.
We know that the complete graph has n(n-1)/2 edges and we want to find out n such that n(n-1)/2 greater or equal to 500. Thus n(n-1) greater or equal to 1000. Taking n=33, we have, n(n-1)=33(33-1)=1056>1000. Therefore required smallest integer is n=33.
well, ace of 1 (or the first number in the equation) =1, d=1 (or what it's added by each time), which shows that it's arithmetic), and n=1000 (the number you're trying to get)the equation is s of n=n/2 (ace of 1 + ace of n) s of n is the sum of the numbersso, s of n=500(1+1000)s of n=500500so 500,500 is the sum of the numbers from 1 to 1000.you can find more athttp://www.youtube.com/watch?v=VgVJrSJxkDk&feature=youtube_gdatahttp://www.youtube.com/watch?v=U_8GRLJplZg&feature=youtube_gdata
1000 m
Sum of first n numbers = n/2(n +1) = 500 x 1001 = 500500
10^(n-1)
All multiples of 3 have digits that add up to a multiple of 3. There are 333 multiples of 3 between 1 and 1000.
To find the sum of 1 to 10, it helps to rearrange them, so instead of thinking of it like: 1 + 2 + . . . + 9 + 10, think like this: (1+10) + (2+9) + (3+8) + (4+7) + (5+6) = 11 + 11 + 11 + 11 + 11 = 5 x 11 = 55.So you have 5, which is 10/2 times 11 which is 10+1.Now to sum 1 to N, it is: (1 + N) + (2 + N-1) + (3 + N-2) + ... = (1+N) + (1+N) + ...., and you do that for N/2 times, so the answer is (1+N)*N/2.So for:N = 100 : 101*100/2 = 5050,N=1000 : 1001*1000/2 = 500500.