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1/3.6+1/6.9+1/9.12+..+1/3n(3n+3)=n/9(n+1)

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1/3.6+1/6.9+1/9.12+..+1/3n(3n+3)=n/9(n+1)

    • Posted by Jay Thacker 7 years, 4 months ago

    CBSE > Class 11 > Mathematics
    • 1 answers

    Suman Sharma 7 years, 4 months ago

    Step 1 put (n=1) L.H.S. 1/3.6=1/18 R.H.S. n/9(n+1)=1/9(1+1) =1/9.2 =1/18 Step 2 Let it be true for (n=k) 1/3.6+1/6.9+1/9.12+...+1/3k(3k+3)=k/9(k+1) Step 3 Put (n=k+1) 1/3.6+1/6.9+1/9.12+...+1/3(k+1)(3(k+1)+3)=k+1/9(k+1+1) =1/3.6+1/6.9+1/9.12+...+1/3k(3k+3)+1/(3k+3)(3k+6)=k+1/3(3k+6) L.H.S. =k/3(3k+3)+1/(3k+3)(3k+6) =1/3k+3(k/3+1/3k+6) =1/3k+3(3k^2+6k+3/3(3k+6)) =1/3k+3((k+1)(3k+3)/3(3k+6)) =1(k+1/3(3k+6)) =k+1/3(3k+6)=R.H.S.

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