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Given that 12 + 22 + 32 + … n2 = n (n + 1) (2n + 1)/6, then 82 + 92 + 102 + …. + 172 is equal to 1). 16162). 16453). 17474). 1555 |
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Answer» GIVEN that, 12 + 22 + 32 + … n2 $(= \;\frac{{n\;\LEFT( {n\; + \;1} \right)\left( {2n\; + \;1} \right)}}{6})$ ⇒ 12 + 22 + 32 + … 172 $(= \;\frac{{17\;\left( {17\; + \;1} \right)\left( {2\; \times 17\; + \;1} \right)}}{6}\; = \;1785)$ And, 12 + 22 + 32 + … 72 $(= \;\frac{{7\;\left( {7\; + \;1} \right)\left( {2\; \times 7\; + \;1} \right)}}{6} = 140)$ HENCE, 82 + 92 + 102 + …. + 172= 1785 – 140 = 1645 |
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