Question

please somebody answer me .​

Answer 1

Answer

A. 12410/21

Step-by-step explanation:

Please note that :

1

1 ----- = 1 + (1/2)

2

So, this series can be divided into two parts :

New series will be :

(1 + 4 + 7 + 10 + ...) +[ (1/2)+(1/6)+(1/12)+(1/20)......]

= 10[2 + (20 -1)3] + (2nd series)

= 590 + (2nd series)

Second series acts like this :

(1/2)+(1/6)+(1/12)+(1/20)......]

=(1/2) + 1/(2+4) + 1/(2+4+6)+ ...

Now, Tn = 1/[2(1+2+3+....)]

= 1/[2*{(n)(n+1)/2}]

= 1/[n(n+1)]

Tn = 1/[(n^2 )+ n]

This is special series .

Now let's decompose it to partial fractions :

1/n(n+1) = (A/n) + B/(n+1)

A(n+1)+B(n)

--------------------- = 1/n(n+1)

n(n+1)

Means, A(n+1) + B(n) = 1

An + Bn +A = 0n

An + Bn = 0n

A= 1 , B= -1

Tn = (1/n) - {1/(n+1)}

So, this is how they will cancel out :

T1 = 1/1 - 1/2

T2 = 1/2 - 1/3

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

Sn = 1 - 1/(n+1)

Means, Sum will be Sn = 1 - 1/(n+1) for 2nd series .

So, S = 590 + (20/21)

= 12410/21