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MISCELLANEOUS EXERCISE - 2 (II) · Q121

Q.If 1+2+3+4+5+⋯ upto n terms1×2+2×3+3×4+4×5+⋯ upto n terms=322\dfrac{1+2+3+4+5+\cdots \text{ upto } n \text{ terms}}{1\times2+2\times3+3\times4+4\times5+\cdots \text{ upto } n \text{ terms}}=\dfrac{3}{22}, find the value of nn.

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Numerator (sum of naturals) =n(n+1)2=\dfrac{n(n+1)}2. Denominator =∑r(r+1)=n(n+1)(n+2)3=\sum r(r+1)=\dfrac{n(n+1)(n+2)}3. Ratio $=\dfrac{n(n+1)/2}{n(n+1)(n+2)/3}=\dfrac3{2(n+2)}=\df …

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