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Q: How many terms are in (r plus 1)?
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What is the r worth in f 1 2 r plus 6.5?

r is worth - 0.542 mathematically if 1 2 r plus 6.5.


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


What is the answer to 1 over R plus 1 over T?

(r+t)/rt


What is the factor of m plus r-rm-1?

There is not the factor: there are two factors: (m - 1) and (1 - r)


Why 1 plus r in NPV calculation?

NPV=NFV/(1+r)^n The role of the "(1+r)^n" is to discount the future money to what it is worth in todays dollars. The 1 accounts to the sum itself and the plus r takes into account the interest rate. NPV=NFV/(1+r)^n The role of the "(1+r)^n" is to discount the future money to what it is worth in todays dollars. The 1 accounts to the sum itself and the plus r takes into account the interest rate.


Can rxr plus 7r plus 6 be factored?

I think so.r2 + 7r + 6 = [ (r+6) (r+1) ]


What is -2(-r -1) plus 1?

It is: 1+9+2 = 12


How do you simplify a plus ar?

( a + ar ) = a ( 1 + r )


What does the 22nd row of Pascal triangle look like?

The terms in row 22 are 22Cr where r = 0, 1, 2, ..., 22. 22Cr = 22!/[r!*(22-r)!] where r! = r*(r-1)*...*3*2*1 and 0! = 1


What is the 29TH row of pascals triangle?

The terms in row 29 are: 29Cr = 29!/[r!*(29-r)!] for r = 0, 1, 2, ... 29 where r! denotes 1*2*3*...*r and 0! = 1


What is the formula for a geometric sequence?

a+a*r+a*r^2+...+a*r^n a = first number r = ratio n = "number of terms"-1


What is 0.5 plus 0.55 plus 0.555 plus .nterms?

Sum of a geometric progression with first term a and constant different r is given by:sum_gp = a(1 - rn)/(1 - r)Required sum is:Sum =0.5 + 0.55 + 0.555 + ... [n terms]= 5 (0.1 + 0.11 + 0.111 + ... [n terms])= 5 (1/10 + 11/100 + 111/1000 + ... [n terms])Multiply by 9/9 (= 1):= 5/9 (9/10 + 99/100 + 999/1000 + ... [n terms])= 5/9 [ (1 - 1/10) + (1 - 1/100) + (1 - 1/1000) + ... + (1 - 1/(10n) ]= 5/9 [ (1 + 1 + 1 + ... [n terms]) - (1/10 + 1/100 + ... + 1/10n) ]= 5/9 [ n - 1/10 (1 - (1/10)n) / (1 - 1/10) ]= 5/9 [ n - 1/9 (1 - (1/10)n) ]= 5/9 [ n - 1/9 (1 - 0.1n) ]= 5/9 [ n - (10n - 1) / (9 x 10n) ]