If `1+2+3+…+k=325`, then find `1^(3)+2^(3)+3^(3)+…+k^(3)`.

Find the value of k if 1^(3) + 2^(3) + 3^(3) +…….. + k^(3) =2025

Find the value of k if 1^(3)+2^(3)+3^(3)+….+k^(3)=2025

If 1^(3) + 2^(3) + 3^(3) + ……. + k^(3) =8281 , then find 1+2 + 3 + ……. +k.

If the system of equation (k+1)^(3)x+(k+2)^(3)y=(k+3)^(3),(k+1)x+(k+2)y=k+3,x+y=1 is consistent,then the value of k is

If 1^(3) + 2^(3) + 3^(3) + . . . . + k^(3) = 44100 then find 1 + 2 + 3 + . . . . + k

If the system of equations (k+1)^(3)x+(k+2)^(3)y=(k+3)^(2),(k+1)x+(k+2)y+k+3,x+y=1 is consistent then the value of k is

If 1^(3)+2^(3)+3^(3)+…+k^(3)=44100 then find 1+2+3+…+k .

The equilibrium constant of the following are : " " {:(N_2+3H_2hArr 2NH_3, K_1),(N_2+O_2hArr 2NO , K_2) , (H_2+(1)/(2) O_2 to H_2O , K_3) :} The equilibrium constant (K) of the reaction : 2NH_3+(5)/(2) O_2 overset(k) hArr 2NO +3H_2O, will be (a) K_1K_3^(3) //K_2 (b) K_2K_3^(3) //K_1 (c) K_2K_3//K_1 (d) K_2^(3) K_3//K_1

If the lines (x-1)/(-3) =(y-2)/(2k) =(z-3)/2 and (x-1)/(3k)=(y-1)/1=(z-6)/(-5) are perpendicular , find the value of k .