Question

What's the sum of the series 1/3×7,1/7×11,1/11×15…up to its nth term?​

Answer 1

Answer:

S= 1/ (3×7)+1/ (7×11)+1/ (11×15)+. ………………….up to nth term. Tn. =1/ (nth term of 3,7,11,15…..) (nth term of 7.11,15,19) Tn = 1/ [3+ (n-1).4]. [7+ (n-1).4] Tn = 1/ (4n-1) (4n+3)=A/ (4n-1)+B/ (4n+3)

Step-by-step explanation:

Answer 2

What's the sum of the series 1/3×7,1/7×11,1/11×15…up to its nth term?

S= 1/(3×7)+1/(7×11)+1/(11×15)+. ………………….up to nth term.

Tn. =1/(nth term of 3,7,11,15…..) ( nth term of 7.11,15,19)

Tn = 1/[3+(n-1).4].[7+(n-1).4]

Tn = 1/(4n-1)(4n+3)=A/(4n-1)+B/(4n+3)

1 =A(4n+3)+B(4n-1)

4A+4B=0………..(1)

3A-B = 1……………(2)

On solving eqn.(1) and (2)

A=1/4. B= -1/4.

Tn= 1/4[ 1/(4n-1) - 1/(4n+3)].

On putting n= 1,2,3,4…………n

T1=1/4[1/3 -1/7]

T2=1/4[1/7–1/11]

T3=1/4[1/11–1/15]

T4=1/4[1/15–1/19]

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Tn=1/4[1/(4n-1) -1/(4n+3)]

On adding :-

T1+T2+T3+T4+………….+Tn = 1/4[1/3–1/(4n+3)].

S= 1/4.[4n+3–3]/3.(4n+3)

S=(1/4).(4n)/3.(4n+3)

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