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MISCELLANEOUS EXERCISE - 2 (II) · Q123

Q.If 1×3+2×5+3×7+⋯ upto n terms12+22+32+⋯ upto n terms=59\dfrac{1\times3+2\times5+3\times7+\cdots \text{ upto } n \text{ terms}}{1^2+2^2+3^2+\cdots \text{ upto } n \text{ terms}}=\dfrac59, find the value of nn.

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Numerator: ∑r(2r+1)=∑(2r2+r)=n(n+1)(2n+1)3+n(n+1)2=n(n+1)(4n+5)6\sum r(2r+1)=\sum(2r^2+r)=\dfrac{n(n+1)(2n+1)}3+\dfrac{n(n+1)}2=\dfrac{n(n+1)(4n+5)}6. Denominator: ∑r3=[n(n+1)2]2\sum r^3=\left[\dfrac{n(n+1)}2\right]^2. Ratio =n(n+1)(4n+5)/6n2(n+1)2/4=2(4n+5)3n(n+1)=59=\dfrac{n(n+1)(4n+5)/6}{n^2(n+1)^2/4}=\dfrac{2(4n+5)}{3n(n+1)}=\dfrac59. Cross-multiplying: $18(4n+5)=15n(n+1) \Rightarrow 72n+90=15n^2+15n \Rightarrow 15n^2-57n- …

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