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Find the sum of n terms of the series (4-1/n)+(4-2/n)+(4-3/n)+............ |
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Answer» Sum=4n-1/n(1+2+3+........n)=4n-1/n*n(n+1)/2=4n-(n+1)/2=1/2(8n-n-1)=1/2(7n-1) Series becomes:4x n-1/n(1+2+3+.......n)=4n-1/n*n(n+1)/2=4n-(n+1)/2=1/2(8n-n-1)={tex}\\frac{7n-1}{2}{/tex} 3n |
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