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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.
