If S_1,S_2,S_3, ... ... ,S_n are the sums of infinite geometric series whose first terms are 1,2,3, ... ... ,n and whose common ratios are (1 / 2),(1 / 3),(1 / 4), ... ... (1 / n + 1) respectively, n then find the values of S_1^2 + S_2^2 + S_3^3 + ... ... .. + S_2n - 1^2.

The sum of the series a+a r+a r^2+ * s ... ... =(a / 1-r) from this the sum of given series =S_k=frack1-(1 / k+1)=k+1 Now the sum of series, here\ S_1^2+S_2^2+S_3^2+ * s ... ... =2^2+3^2+ * s ... +(2 n)^2 =1^2+2^2+3^2+ * s ... ... (2 n)^2-1^2 \ =((2 n)(2 n+1)(2 x 2 n+1) / 6)-1 =(n(2 n+1)(4 n+1)-3 / 3)

If S_{1},S_{2},S_{3},,S_{n} are the sums of infinite ... | Sequence & Series (Mathematics)

The sum of the series $a+a r+a r^2+\cdots \ldots \ldots=\frac{a}{1-r}$ from this the sum of given series $=S_k=\frac{k}{1-\frac{1}{k+1}}=k+1\\$ Now the sum of series, here $\\ S_1^2+S_2^2+S_3^2+\cdots \ldots \ldots$ $\\=2^2+3^2+\cdots \ldots+(2 n)^2$ $\\=1^2+2^2+3^2+\cdots \ldots \ldots(2 n)^2-1^2$ $\\ =\frac{(2 n)(2 n+1)(2 \times 2 n+1)}{6}-1$ $\\=\frac{n(2 n+1)(4 n+1)-3}{3}$

Explanation

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