`sum_(r=1)^(10) (4r-3)^(2)` =

sum_(r=1)^(10)(r)/(1-3r^(2)+r^(4))

If sum_(r=1)^(10)(r.3^(r))=(19.3^(a)+b)/(4) , then

If sum_(r=1)^(10)r(r-1)10C_(r)=k.2^(9), then k is equal to- -1

Sum the following series to n terms and to infinity : (i) (1)/(1.4.7) + (1)/(4.7.10) + (1)/(7.10.13) + ... (ii) sum_(r=1)^(n) r (r + 1) (r + 2) (r + 3) (iii) sum_(r=1)^(n) (1)/(4r^(2) - 1)

sum_(r=0)^(10) (-1)^(r) (""^(10)C_(r))/(""^(r+2)C_(r)) is equal to .

Let sum_(r=1)^(n)r^(4)=f(n)," then " sum_(r=1)^(n) (2r-1)^(4) is equal to

If C_(r)=10C_(r), then sum_(r=1)^(10)C_(r-1)C_(r) is equal to

sum_(r=1)^10 r/(1-3r^2+r^4)

Find sum_(r=1)^(10) (r+2)(3r+1).