Question

plz give me detailed solution​

Answer 1

the series is

1/(1x4) + 1/(4x7) + 1/(7x10) + ..........+ 1/(16x19)

the nth term.of the series can be written as

Tn = 1/(3n - 2)(3n + 1)

now, we will represent the term as addition/subtraction of two different terms

let

A/(3n-2) + B/(3n+1) = 1/(3n-2)(3n+1)

A(3n+1) + B(3n-2) = 1.

(3n-2)(3n+1). (3n-2)(3n+1)

comparing numerator

A(3n+1) + B(3n-2) = 1

3An + A + 3Bn - 2B = 1

3n(A + B) + A - 2B = 1

since RHS does not contain n so

A + B = 0

A - 2B = 1

subtracting

3B = -1

B = -1/3

A = -B = -(-1/3) = 1/3

thus

Tn = (1/3) 1/((3n-2) - (1/3) 1/(3n+1)

putting n = 1,2,3...

T1 = (1/3) [ 1/1 - 1/4]

T2 = (1/3) [ 1/4 - 1/7]

T3 = (1/3) [ 1/7 - 1/10]

..................................

.................................

T6 = (1/3) [ 1/16 - 1/19]

adding the above terms

T1+t2+T3+....+T6 = (1/3) [1 - 1/19]

( all other terms in RHS would cancel out)

Sum of series = (1/3)*18/19

= 6/19