Sum the series `1+(1+a)r+(1+a+a^2)r^2+……`upto `infty`, where |a|<1,|r|<1.

Sum of the series sum_(r=1)^(n) (r^(2)+1)r! is

The sum of the series 1+3x+6x^2+10x^3+………….infty is:

The value of 9^(1/3), 9^(1/9).9^(1/27)…………..upto infty is:

The sum of the series is 1+4/5+7/5^2+10/5^3+…………infty is:

Consider the binomial expansion of R = (1 + 2x )^(n) = I + f , where I is the integral part of R and f is the fractional part of R , n in N . Also , the sum of coefficient of R is 2187. The value of (n+ Rf ) "for x" = (1)/(sqrt(2)) is

Find the sum of the series sum_(r=1)^(n) r x^(r-1) using calculus .

If the coefficients of a^(r-1), a^(r ), a^(r+1) in the binomial expansion of (1 + a)^(n) are in Arithmetic Progression, prove that: n^(2) -n(4r+1) + 4r^(2) -2=0

A galvanometer of resistance G is converted into a voltmeter to measure upto V volts by connecting a resistance R_1 in series with the coil. If a resistance R_2 s connected in series with it then it can measure upto V/2 volts Find the resistance R, in terms of R_1 and R_2 required to convert it into a voltmeter that can read upto 2V volts.

Let V_(r ) denotes the sum of the first r terms of an arithmetic progression whose first term is r and the common difference is (2r-1) . Let T_(r )=V_(r+1)-V_(r)-2" and " Q_(r )=T_(r+1)-T_(r )" for " r=1,2,"...." The sum V1​+V2​+...+Vn​ is