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.

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))

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

Prove that sum_(r=1)^(n)(-1)^(r-1)(1+(1)/(2)+(1)/(3)+...+(1)/(r))^(n)C_(r)=(1)/(n)

A network of resistance is constructed with R_1 and R_2 as shown in fig. The potential at the points 1,2,3 …., N are V_1, V_2,V_3,…,V_n, respectively, each having a potential k times smaller then the previous one. The ratio R_1//R_2 is

Ifn>2than sum_(r=0)^(n)(-1)^(r)(n-r)(n-r+1)C_(r)=(A)0(B)n(C)2^(n)(D)(n-1)2^(n)

The value of sum_(r=0)^(3n-1)(-1)^(r)6nC_(2r+1)3^(r) is