Question

the sum of n+1 terms of 1/1+1/1+2+1/1+2+3 is​

Answer 1

Step-by-step explanation:

1/1+1/(1+2)+1/(1+2+3(+………..+1/(1+2+3+….n)

tn =1/sigma n = 2/n.(n+1)

Now 2/n.(n+1) = A/n + B/(n+1)

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

By compairing both sides

A+B = 0…………(1)

A=2………………….(2)

2+B = 0 => B= -2

tn = 2/n -2/(n+1).

On putting n= 1,2,3,……n.

t1 =2/1 - 2/2

t2=2/2 - 2/3

t3 = 2/3 -2/4

××××××××××××

××××××××÷÷÷÷

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

On adding

.t1+t2+t3+…………….+tn = 2/1 -2/(n+1)

= 2.(n+1–1)/(n+1) = 2.n/(n+1) .Answer.

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