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How do you find the sum of an infinite geometric series?
An infinite geometric series can be summed only if the common ratio has an absolute value less than 1. Suppose the sum to n terms is S(n). That is, S(n) = a + ar + ar2 + ... + arn-1 Multipying through by the common ratio, r, gives r*S(n) = ar + ar2 + ar3 + ... + arn Subtracting the second equation from the first, S(n) - r*S(n) = a - arn (1 - r)*S(n) = a*(1 - rn) Dividing by (1 - r), S(n) = (1 - rn)/(1 - r) Now, since |r| < 1, rn tends to 0 as n tends to infinity and so S(n) tends to 1/(1 - r) or, the infinite sum is 1/(1 - r)
With this radical n equals S r exponent-2 solve for r?
sqrt(n) = S*r2sqrt(n)/S = r2 sqrt(sqrt(n)/S) = ror4th root of n/sqrt(S) = r.If your equation was sqrt(n) = S*r-2sqrt(n) = S*(1/r2)sqrt(n)/S = 1/r2S/sqrt(n) = r2 sqrt(S/sqrt[n]) = rorsqrt(S)/4th root of n = r
Prove that nCr plus nCr minus 1 equals n plus 1Cr?
nCr + nCr-1 = n!/[r!(n-r)!] + n!/[(r-1)!(n-r+1)!] = n!/[(r-1)!(n-r)!]*{1/r + 1/n-r+1} = n!/[(r-1)!(n-r)!]*{[(n-r+1) + r]/[r*(n-r+1)]} = n!/[(r-1)!(n-r)!]*{(n+1)/r*(n-r+1)]} = (n+1)!/[r!(n+1-r)!] = n+1Cr
Factor this expression mr ns - nr - ms?
I am guessing there is a missing plus sign and you want to factor mr + ns - nr - ms. If so , mr -ms + ns - nr = m(r - s) - n( r -s ) = (r - s) (m - n)
1 plus r raised to power n equals?
Binomial Theorem: 1n + nC1*1n-1*r + nC2*1n-2*r2+......+nCn-1*1*rn-1 + rn Or (1+r)n = 1 + n*r + n(n-1)/2! * r2 + n(n-1)(n-2)/3! * r3 + .......... n(n-1)...(n-k)/k! * rk if n < 1 as you cannot calculate the combinations that easily. This gives an accurate approximation provided that abs(x) < 1.
How do you find the sum of an infinite geometric series?
An infinite geometric series can be summed only if the common ratio has an absolute value less than 1. Suppose the sum to n terms is S(n). That is, S(n) = a + ar + ar2 + ... + arn-1 Multipying through by the common ratio, r, gives r*S(n) = ar + ar2 + ar3 + ... + arn Subtracting the second equation from the first, S(n) - r*S(n) = a - arn (1 - r)*S(n) = a*(1 - rn) Dividing by (1 - r), S(n) = (1 - rn)/(1 - r) Now, since |r| < 1, rn tends to 0 as n tends to infinity and so S(n) tends to 1/(1 - r) or, the infinite sum is 1/(1 - r)
The S R of -1 is an I N?
the question is = The S R of -1 is an I N = so the answer is: the square root of -1 is an imaginary number
With this radical n equals S r exponent-2 solve for r?
sqrt(n) = S*r2sqrt(n)/S = r2 sqrt(sqrt(n)/S) = ror4th root of n/sqrt(S) = r.If your equation was sqrt(n) = S*r-2sqrt(n) = S*(1/r2)sqrt(n)/S = 1/r2S/sqrt(n) = r2 sqrt(S/sqrt[n]) = rorsqrt(S)/4th root of n = r
Prove that nCr plus nCr minus 1 equals n plus 1Cr?
nCr + nCr-1 = n!/[r!(n-r)!] + n!/[(r-1)!(n-r+1)!] = n!/[(r-1)!(n-r)!]*{1/r + 1/n-r+1} = n!/[(r-1)!(n-r)!]*{[(n-r+1) + r]/[r*(n-r+1)]} = n!/[(r-1)!(n-r)!]*{(n+1)/r*(n-r+1)]} = (n+1)!/[r!(n+1-r)!] = n+1Cr
Implimentation of pascle triangle in C?
#include#includevoid main(){int n,a=1,s=1,r;printf("\n enter number of lines ");cin>>n;for(;a=1;b--){printf(" ");}r=pow(s,2);printf("\n");a++;}getch();}// \m/
The S R of -1 is an 1 N?
the square root of -1 is an imaginary number
Why is it said that poisson distribution is a limiting case of binomial distribution?
This browser is totally bloody useless for mathematical display but...The probability function of the binomial distribution is P(X = r) = (nCr)*p^r*(1-p)^(n-r) where nCr =n!/[r!(n-r)!]Let n -> infinity while np = L, a constant, so that p = L/nthenP(X = r) = lim as n -> infinity of n*(n-1)*...*(n-k+1)/r! * (L/n)^r * (1 - L/n)^(n-r)= lim as n -> infinity of {n^r - O[(n)^(k-1)]}/r! * (L^r/n^r) * (1 - L/n)^(n-r)= lim as n -> infinity of 1/r! * (L^r) * (1 - L/n)^(n-r) (cancelling out n^r and removing O(n)^(r-1) as being insignificantly smaller than the denominator, n^r)= lim as n -> infinity of (L^r) / r! * (1 - L/n)^(n-r)Now lim n -> infinity of (1 - L/n)^n = e^(-L)and lim n -> infinity of (1 - L/n)^r = lim (1 - 0)^r = 1lim as n -> infinity of (1 - L/n)^(n-r) = e^(-L)So P(X = r) = L^r * e^(-L)/r! which is the probability function of the Poisson distribution with parameter L.
Write a program defind the sum of dijit of a number?
Void main () { Int n, r, s = 0 ; Printf (" Enter the number") Scanf ("% d", & n); While (n! = 0) { r = n%/10; s = s+r; n = n/10; s = s+r; n = n/10 } Printf ("sun of the dijit of a no. is % d; s)
How do you spell sirens?
Sirens is spelled S-I-R-E-N-S.
What is the difference between an arithmetic series and a geometric series?
An arithmetic series is the sequence of partial sums of an arithmetic sequence. That is, if A = {a, a+d, a+2d, ..., a+(n-1)d, ... } then the terms of the arithmetic series, S(n), are the sums of the first n terms and S(n) = n/2*[2a + (n-1)d]. Arithmetic series can never converge.A geometric series is the sequence of partial sums of a geometric sequence. That is, if G = {a, ar, ar^2, ..., ar^(n-1), ... } then the terms of the geometric series, T(n), are the sums of the first n terms and T(n) = a*(1 - r^n)/(1 - r). If |r| < 1 then T(n) tends to 1/(1 - r) as n tends to infinity.
What is the general term of the expansion 1-x to the power n?
The r+1 th term is nCr(-x)r where r = 0, 1, 2, ... , n. and where nCr = n!/[r!*(n-r)!]
Write a C program to convert decimal number into binary number?
#include<iostream> using namespace std; int revDigit(int n){ int r,s=0; while(n>0){ r=n%10; n=n/10; s=(s*10)+r; } return s; } main(){ long n,r,r1,s=0,s1=10; cout<<"Enter a number."<<endl; cin>>n; while(n>0){ r=n%2; n=n/2; s=(s*10)+r; } //since we read the binary equivalent in a bottom up manner, we need to reverse it. s1=revDigit(s); cout<<"Binary equivalent:\t"<<s1<<endl; system("pause");//this line is required if you are using dev c++ }
