Result 2 If `V_0` is the value of an article at a certain time and the rate of depreciation is `R_1 %` for first `n_1` years `r_2 %` for next `n_2` years and so on and `R_k %` for the last `n_k` years then the `V=V_0(1- R_1/100)^(n1) (1- R_2/100)^(n2) .......... (1- R_3/100)^(nk)

Ressult 1 If V_(0) is the value of an article at a certain time and R% per annum is the rate of depreciation then the value V_(n) at the end of n years is given by V_(n)=V_(0)(1-(R)/(1000^(n))

Let P be the principal and the rate of interest be R_1 % for first year R_2 % for second year R_3 % for third year and so on and in the last R_n % for the nth year. Then the amount A and the compound interest C.I. at the end of n years are given by A=P(1+R_1/100) (1+ R_2/100) ...... (1+ R_n/100) and C.I. =A-P respectively.

If the principal be Rs. p and the annual rate of interest be r_(1)% in the 1st year r_(2)% in the 2nd years, then write the amount after n years.

Formula 2 Let P be the population of a city or a town at the beginning of a certain year.If the population grows at the rate of R_(1)% during first year and R_(2) during second year then population after 2 years =P(1+(R_(1))/(100))xx(1+(R_(2))/(100))

A. Write true of false: a. If Principal =Rs.p , rate of compound interest per annum =r% and time =n years, then the amount after 2 years is Rs. p(1+r/100)^(2) b. If principal =Rs. p and rate of compound interest per annum for first, 2nd and 3rd year be r_(1)%, r_(2)% and r_(3)% respectively, then total amount for 3 years. =Rs. p(1+(r_(1))/100)^(3)+Rs. p(1+(r_(2))/100)^(3)+Rs. p(1+(r_(3))/100)^(3)

The annual rate of compound interest in 1st year be r_1% , in 2nd year be r_2% and the 3rd year be r_3% , then the total amount of principal ₹ p in 3 years =

Formula 2 Let P be the principal and the rate of interest be R% per annum.If the interest is compounded annually then the amount A and the compound interest C.I.at the end of n years are given by A=P(1+(R)/(100k))^(nk) and C.I.A-P=P{((1_(R))/(100k))^(nk)-1} respectively

For integer n gt 1 , the digit at units place in the number sum_(r=0)^(100)r!+2^(2^n) is

If sum_(r = 0)^(2n) a_(r ) (x-2)^(r ) = sum_(r = 0)^(2n) b_(r ) (x-3)^(r ) and a_(k ) =1 for all k ge n then show that b_(n ) =""^((2n+1)) C_((n+1)) .