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Question

If S1,S2,S3,,Sn are the sums of infinite G.P.s. whose first terms are 1,2,3,,n and whose common ratios are 12,13,14,...,1n+1 respectively, then i=1nSi=


A

nn+32

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B

nn+42

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C

nn-32

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D

nn+12

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Solution

The correct option is A

nn+32


Explanation for the correct option

Step 1:Solve using sum of infinite terms of GP

We know that the formula for the sum of the G.P with infinite terms is given by,

S=a1-r, where a is the initial term and r is the common ratio.

Then the given summation can be written as,

i=1nSi==i=1na1-r

It is given that initial terms and common ratio of infinite G.P's are 1,2,3,,n and 12,13,14,...,1n+1 respectively.

From this, we can find that

S1=11-12=2

S2 will be,

S2=21-13=3

S3 can be found as

S3=31-14=4

Similarly, Sn will be

Sn=n1-1n+1=n+1

Step 2: Calculation for the required summation

With the help of the above values, we can write the i=1nSi as

i=1nSi=S1+S2+S3++Sn=2+3+4++n+1

We can see that on the RHS there is an A.P with an initial term a=2, common difference d=1, and nth term is n.

Now, the formula of the sum of an A.P is given by

Sn=n2(2a+n-1d)

Then, the sum will be

Sn=n222+n-11=n23+n

With the help of this, we get

i=1nSi=n3+n2

Hence, the correct option is (A).


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