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The nth term in the sum is t_n=[n(n+1)(2n+1)/6]^2/[n(n+1)/2]^2= (2n+1)^2/9. So, the required sum is the sum of t_n= 4/9 n^2 + 4/9 n +1/9 from n=1 to n=10, which gives S=590/3.
590/3
4/6+{4+6}/{3+6}={4/6+10/9}=14/15+(1^2+3^2)^2/1^3+3^3=14/15+{4+12}/{3+9}={14/15+16/12}=30/27+(1^2+4^2)^2/1^3+4^3=30/27+{4+16}/{3+12}={30/27+20/15}=50/42+(1^2+5^2)^2/1^3+5^3=50/42+{4+20}/{3+15}={50/42+24/18}=74/60+(1^210^2)^2/1^3+10^3=74/60+{4+40}/{3+30}={74/60+44/33}=118/93=18.4 2^9.4 2^3^2.2^2 13^1.1^2 3.2 (n ➖ 3n+2).
The nth term in the sum is t_n=[n(n+1)(2n+1)/6]^2/[n(n+1)/2]^2= (2n+1)^2/9. So, the required sum is the sum of t_n= 4/9 n^2 + 4/9 n +1/9 from n=1 to n=10, which gives S=590/3.
590/3
4/6+{4+6}/{3+6}={4/6+10/9}=14/15+(1^2+3^2)^2/1^3+3^3=14/15+{4+12}/{3+9}={14/15+16/12}=30/27+(1^2+4^2)^2/1^3+4^3=30/27+{4+16}/{3+12}={30/27+20/15}=50/42+(1^2+5^2)^2/1^3+5^3=50/42+{4+20}/{3+15}={50/42+24/18}=74/60+(1^210^2)^2/1^3+10^3=74/60+{4+40}/{3+30}={74/60+44/33}=118/93=18.4 2^9.4 2^3^2.2^2 13^1.1^2 3.2 (n ➖ 3n+2).