Question

sum the series √3+3√2+6√3+.........to 16terms
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Answer 1

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Step-by-step explanation:

this series √3+3√2+6√3+......... is in a GP where,

a = √3 , n = 16 , r = 3√2/√3 = √6

sum of series, S = a(r^n - 1)/(r - 1)

S = √3((√6)^16 - 1)/(√6 - 1)

by applying a^2-b^2= (a+b)(a-b) continuously,

S = √3 (√6^8+1)(√6^4+1)(√6^2+1)(√6+1)(√6-1) / (√6-1)

S = √3 (6^4+1)(6^2+1)(6+1)(√6+1)

S = √3 (1296+1)(36+1)(6+1)(√6+1)

S= √3 × 1297 × 37 × 7 × (√6+1)

S = √3 (√6+1) × 335923

S = 335923(3√2 + √3) or (1007769√2 + 335923√3) or 1946860.21