Question

Give proving of an=Sn-Sn-1

Answer 1

Answer:

The idea lies at the definition of Sn.

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Suppose a series: a1, a2, a3, a4, a5, ..., a(n-1), an, a(n+1), ....

Here, a1= first term,

an= nth term of the series,

Sn= sum of first n terms of the series,

S(n-1)= sum of first (n-1) terms of the series.

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Therefore, S1= a1 = sum of very first term only

S2= a1+ a2 = sum of first two terms

S3= a1+ a2+ a3 = sum of first three terms

...

S(n-1)= a1+ a2+ a3+ ...+ a(n-1)

Sn= a1+ a2+ a3+ ...+ a(n-1)+ an

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Therefore,

Sn= [ a1+ a2+ a3+ ...+ a(n-1) ]+ an

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Sn= S(n-1) + an

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On substracting 'S(n-1)' on both the sides, we get,

Sn- S(n-1)= an.

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For example =>

assume a series: 4,6,8,3,5,2

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Here n=6= number of last term

a1=4 = first term

a6=2 = an= last term

S1= a1= 4

S2= 4+6

S3= 4+6+8

S4= 4+6+8+3

S5= 4+6+8+3+5

S6= 4+6+8+3+5+2

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Thus, S6= (4+6+8+3+5) +2

S6= S5+ 2

Therefore, S6 - S5=2 -------- ( i )

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Now put n=6 here in equation ( i ), we get,

Sn- S(n-1) = 2 ----- ( ii )

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Here n=6 and (n-1)=5.

Also since, a6= 2= an= last term.

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Hence from ( ii ),

Sn- S(n-1)= an.