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Exercise 5.5 · Q15

Q.The sum up to nn terms of the series 2+8+18+32+⋯\sqrt2+\sqrt8+\sqrt{18}+\sqrt{32}+\cdots is

(1) n(n+1)2\dfrac{n(n+1)}{2}
(2) 2n(n+1)2n(n+1)
(3) n(n+1)2\dfrac{n(n+1)}{2}
(4) 11
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Simplify each radical to a multiple of 2\sqrt2 to reveal the underlying AP 1,2,3,…1,2,3,\ldots, then sum.

Step 1. Simplify each term. 2=1⋅2\sqrt2=1\cdot\sqrt2,  8=4×2=22\ \sqrt8=\sqrt{4\times2}=2\sqrt2,  18=9×2=32\ \sqrt{18}=\sqrt{9\times2}=3\sqrt2,  32=16×2=42\ \sqrt{32}=\sqrt{16\times2}=4\sqrt2 — the kthk^{th} term is k2k\sqrt2.

Step 2. Sum to nn terms.

Sn=2(1+2+⋯+n)=2⋅n(n+1)2=2 n(n+1)2.S_n = \sqrt2(1+2+\cdots+n) = \sqrt2\cdot\frac{n(n+1)}2 = \frac{\sqrt2\,n(n+1)}2. …

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