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Question

51·2·3+73·4·5+95·6·7+... is equal to


A

log8e

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B

log8e

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C

loge8

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D

none of these

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Solution

The correct option is A

log8e


Explanation for the correct option:

Step 1: Find the nth term of the series:

Given: 51·2·3+73·4·5+95·6·7+...

It can also be written as 12n+3(2n-1)2n(2n+1).

Step 2: Break the obtained expression into partial expressions:

Simplify the expression by breaking it into smaller fractions.

2n+3(2n-1)2n(2n+1)=A2n-1+B2n+C(2n+1)

2n+3(2n-1)2n(2n+1)=2n(2n+1)A+B2n-1(2n+1)+2n(2n-1)C2n-12n(2n+1)

2n+3=2n(2n+1)A+B2n-1(2n+1)+2n(2n-1)C

2n+3=4n2A+2nA+4n2B-B+4n2C-2nC

Comparing coefficient of n2,n and constant

4n2A+B+C=0n2, -B=3, 2n(A-C)=2n

A=2,B=-3,C=1

12n+3(2n-1)2n(2n+1)=122n-1-32n+12n+1=122n-1-22n+112n+1-12n=2112n-1-12n+112n+1-12n=2112n2n-1+112n2n+1

Step 3: Use logarithmic equations to get the summation.

2112n2n-1+112n2n+1=21-12+13....+13-12+15-14=2log2+(log2-1)=3log2-1=log8-loge=log(8e) log(1-x)=x-x2+x3.....

Hence, option (A) is the correct answer.


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