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Question

Let Sn(1n9) denote the sum of n terms of the series 1+22+333+...+999...99times then for 2n9

A
SnSn1=19(10nn2+n)
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B
Sn=19(10nn2+2n2)
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C
9(SnSn1)=n(10n1)
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D
none of these
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Solution

The correct option is C 9(SnSn1)=n(10n1)
Given Sn=1+22+333+...+999...9(9times)for(1n9)
For n=9,
Sn=1+22+333+...+888...8(8times)+999...9(9times)
Therefore, Sn1=1+22+333+...+888...8(8times)
Consider, (SnSn1)=(1+22+333+...+888...8(8times)+999...9(9times))(1+22+333+...+888...8(8times))
(SnSn1)=999...9
(SnSn1)=9(111...1)
(SnSn1)=9(108+107+106+...+100)
(SnSn1)=9(10n1)9
9(SnSn1)=9(10n1)
9(SnSn1)=n(10n1)....(sincen=9)
Similarly, for n=8,
Sn=1+22+333+...+888...8(8times)
Sn=1+22+333+...+777...7(7times)+888...8(8times)
Therefore, Sn1=1+22+333+...+777...7(7times)
Consider, (SnSn1)=(1+22+333+...+777...7(7times)+888...8(8times))(1+22+333+...+777...7(7times))
(SnSn1)=888...8
(SnSn1)=8(111...1)
(SnSn1)=8(107+106+105+...+100)
(SnSn1)=8(10n1)9
9(SnSn1)=8(10n1)
9(SnSn1)=n(10n1)....(sincen=8)
In general, 9(SnSn1)=n(10n1)
Thus, 9(SnSn1)=n(10n1), for(2n9).

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