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  • Find the sum to n terms of each of the series in 5^2 + 6^2 + 7^2 + ... + 20^2

5.   Find the sum to n terms of each of the series in 5 ^ 2 + 6 ^ 2 + 7 ^ 2 + ....+ 2 0 ^2

Answers (1)

best_answer

series =    5 ^ 2 + 6 ^ 2 + 7 ^ 2 + ....+ 2 0 ^2

n th term  = (n+4)^2=n^2+8n+16=a_n

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} (k+4)^2

                        =\sum _{k=1}^{n} k^2+8\sum _{k=1}^{n} k+\sum _{k=1}^{n}16

                     =\frac{n(n+1)(2n+1)}{6}+\frac{8.n(n+1)}{2}+16n

           16th term is (16+4)^2=20^2

S_1_6=\frac{16(16+1)(2(16)+1)}{6}+\frac{8.(16)(16+1)}{2}+16(16)

          S_1_6=\frac{16(17)(33)}{6}+\frac{8.(16)(17)}{2}+16(16)

        S_1_6=1496+1088+256

    S_1_6=2840

Hence, the sum of the series 5 ^ 2 + 6 ^ 2 + 7 ^ 2 + ....+ 2 0 ^2  is 2840.

Posted by

seema garhwal

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