(n)/(4)-5=(n)/(6)+(1)/(2)}

(1)/(2.5)+(1)/(5.8)+(1)/(8.11)+... n terms =(A)(n)/(6n+4) (B) (n)/(3n+2)(C)(n)/(4n+6)(D)(1)/(2(2n+3))

n if (1)/(4l)+(1)/(5!)+(1)/(6!)=(n)/(7!)

(1^(4))/(1.3)+(2^(4))/(3.5)+(3^(4))/(5.7)+......+(n^(4)) /((2n-1)(2n+1))=(n(4n^(2)+6n+5))/(48)+(n)/(16(2n+1))

In the binomial expansion of (a-b)^(n),n>=5 the sum of 5 th and 6 th terms is zero,then (a)/(b) equals (1)(5)/(n-4) (2) (6)/(n-5) (3) (n-5)/(6) (4) (n-4)/(5)

Prove the following by using the principle of mathematical induction for all n in Nvdots(1)/(2.5)+(1)/(5.8)+(1)/(8.11)+...+(1)/((3n-1)(3n+2))=(n)/((6n+4))=(n)/((6n+4))

tan^(-1)((n-5)/(n-6))+tan^(-1)((n+5)/(n+6))=(pi)/(4)

1+(n)/(2)+(n(n-1))/(2.4)+(n(n-1) (n-2))/(2.4.6)+…....=

The sum of the series 1+4+3+6+5+8+ upto n term when n is an even number (n^(2)+n)/(4) 2.(n^(2)+3n)/(2) 3.(n^(2)+1)/(4) 4.(n(n-1))/(4)(n^(2)+3n)/(4)

Prove that by using the principle of mathematical induction for all n in N : (1)/(2.5)+ (1)/(5.8) + (1)/(8.11)+ ...+(1)/((3n-1)(3n+2))= (n)/(6n+4)

Prove that by using the principle of mathematical induction for all n in N : (1)/(2.5)+ (1)/(5.8) + (1)/(8.11)+ ...+(1)/((3n-1)(3n+2))= (n)/(6n+4)