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सिध्द कीजिए कि `(1xx2^(2)+2xx3^(2)+..+n(n+1)^(2))/(1^(2)xx2+2^(2)xx3+...+n^(2)(n+1))=(3n+5)/(3n+1)` |
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Answer» माना अंश (Numerator)का nवाँ पद `T_(n)`हैं | `:. " "T_(n)=n(n+1)^(2)=n^(3)+2n^(2)+n` `:. " "अंश =SigmaT_(n)=Sigma(n^(3)+2n^(2)+n)=Sigman^(3)+2Sigma n^(2)+Sigma n` `=(n^(2)(n+1)^(2))/(4)+2((n(n+1)(2n+1))/(6))+(n(n+1))/(2)` ` =(n(n+1))/(12)[3n(n+1)+4(2n+1)+6]` `=(n(n+1)(3n^(2)+11n+10))/(12)` `=(n(n+1)(n+2)(3n+5))/(12)" "`...(i) इस प्रकार हर (denominator)का nवाँ पद `=n^(2)(n+1)=n^(3)+n^(2)` `:. " " हर =Sigma n^(3)+Sigma n^(2)` `=(n^(2)(n+1)^(2))/(4)+(n(n+1)(2n+1))/(6)=(n(n+1))/(12)[3n(n+1)+2(2n+1)]` `=(n(n+1)(3n^(2)+7n+2))/(12)` `=(n(n+1)(n+2)(3n+1))/(12)" "`...(ii) समीकरण (i) व (ii)से `(1xx2^(2)+2xx3^(2)+...+n(n+1)^(2))/(1^(2)xx2+2^(2)xx3+..+n^(2)(n+1))=(3n+5)/(3n+1)""`यही सिध्द करना था | |
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