Question

find the value of given series ​

Answer 1

1/{(ar+b)(ar+a+b)}=1/a{a/(ar+b)(ar+a+b)

={(ar+a+b)-(ar+b)}/{(ar+b)(ar+a+b)×a}

=(1/a){1/(ar+b) -1/(ar+a+b)}

putting r=1 to r=n

we get

=(1/a){1/(a+b) -1/(2a+b) +1/(2a+b)-1/(3a+b)+1/(3a+b)-(1/4a+b)...........+1/(an+b)-1/(an+a+b)}

clearly we can see that, consecutive terms will be cancelled out and finally (1/a)(1/(a+b)-1/(an+a+b) is left

so,the sum=(1/a){(1/(a+b) -1/(an+a+b)}

=(1/a){(an+a+b-a-b)/(an+a+b)(a+b)

=n/{a(n+1)+b}(a+b)

press the like option,,if u understand it

Answer 2

Answer:

n/{a(n+1)+b}(a+b)

Step-by-step explanation:

1/{(ar+b)(ar+a+b)}=1/a{a/(ar+b)(ar+a+b)

={(ar+a+b)-(ar+b)}/{(ar+b)(ar+a+b)×a}

=(1/a){1/(ar+b) -1/(ar+a+b)}

putting r=1 to r=n

we get

=(1/a){1/(a+b) -1/(2a+b) +1/(2a+b)-1/(3a+b)+1/(3a+b)-(1/4a+b)...........+1/(an+b)-1/(an+a+b)}

clearly we can see that, consecutive terms will be cancelled out and finally (1/a)(1/(a+b)-1/(an+a+b) is left

so,the sum=(1/a){(1/(a+b) -1/(an+a+b)}

=(1/a){(an+a+b-a-b)/(an+a+b)(a+b)

=n/{a(n+1)+b}(a+b)