Find the sum of n terms of the series (4-1/n)+(4-2/n)+(4-3/n)
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sum of 4-1/n, 4-2/n, 4-3/n up to the nth term
= (4 + 4 + 4 + 4 + 4 + ......... upto n terms) + (-1/n - 2/n - 3/n - ..........upto n terms)
= 4 ( 1+1+1+1.......... upto n terms) - 1/n (1 + 2 + 3 +4 .........upto n terms)
= 4 n - 1/n × n(n +1)/2
= 4n - (n+1)/2
= [ 8n - (n+1) ] / 2 .......taking L.C.M
=( 7n - 1) / 2 Answer
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= (4 + 4 + 4 + 4 + 4 + ......... upto n terms) + (-1/n - 2/n - 3/n - ..........upto n terms)
= 4 ( 1+1+1+1.......... upto n terms) - 1/n (1 + 2 + 3 +4 .........upto n terms)
= 4 n - 1/n × n(n +1)/2
= 4n - (n+1)/2
= [ 8n - (n+1) ] / 2 .......taking L.C.M
=( 7n - 1) / 2 Answer
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Answered by
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Answer:
( 7n - 1) / 2
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