Question

Good morning!!
Can anyone pls answer this!!
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Answer 1

Hii...

a1+a2+a3+at+at=2(√a1-1+2√a2-4+3√a3-9+4√a4-16+5√a5-25

Answer 2

$\mathsf{Let\;us\;Consider\;the\;LHS\;of\;the\;Given\;Equation :}$

✿ $\mathsf{1\sqrt{a_1 - 1} + 2\sqrt{a_2 - 4} + 3\sqrt{a_3 - 9} + 4\sqrt{a_4 - 16} + 5\sqrt{a_5 - 25}}$

$\mathsf{We\;can\;Write\;the\;Given\;LHS\;as\;\sum (n = 1)\;n\sqrt{a_n - n^2}}$

$\mathsf{Let\;us\;Consider\;the\;RHS\;of\;the\;Given\;Equation :}$

✿ $\mathsf{(\dfrac{a_1 + a_2 + a_3 + a_4 + a_5}{2})}$

$\mathsf{We\;can\;Write\;the\;Given\;RHS\;as\;\sum(n = 1)(\dfrac{a_n}{2})}$

$\mathsf{\implies The\;Given\;Question\;Changes\;to :}$

✿ $\mathsf{\sum(n = 1)\;n\sqrt{a_n - n^2} = \sum(n = 1)(\dfrac{a_n}{2})}$

$\mathsf{Now,\;Consider\;the\;Situation : n = 1}$

$\mathsf{\implies \sqrt{a_1 - 1} = \dfrac{a_1}{2}}$

$\mathsf{Squaring\;on\;both\;Sides,\;We\;get :}$

$\mathsf{\implies (a_1 - 1) = \dfrac{(a_1)^2}{4}}$

$\mathsf{\implies (a_1)^2 - 4(a_1) + 4 = 0}\\\\\mathsf{\implies (a_1 - 2)^2 = 0}\\\\\mathsf{\implies a_1 = 2}$

$\mathsf{Now,\;Consider\;the\;Situation : n = 2}$

$\mathsf{\implies \sqrt{a_1 - 1} + 2\sqrt{a_2 - 4} = (\dfrac{a_1 + a_2}{2})}\\\\\mathsf{\implies \sqrt{2 - 1} + 2\sqrt{a_2 - 4} = (\dfrac{2 + a_2}{2})}\\\\\mathsf{\implies 1 + 2\sqrt{a_2 - 4} = (1 + \dfrac{a_2}{2})}\\\\\mathsf{\implies 2\sqrt{a_2 - 4} = (\dfrac{a_2}{2})}\\\\\mathsf{Squaring\;on\;both\;Sides,\;We\;get:}$

$\mathsf{\implies 16(a_2 - 4) = (a_2)^2}\\\\\mathsf{\implies (a_2)^2 - 16(a_2) + 64}\\\\\mathsf{\implies (a_2 - 8)^2 = 0}\\\\\mathsf{\implies a_2 = 8}$

$\mathsf{Now,\;Consider\;the\;Situation : n = 3}$

$\mathsf{\implies \sqrt{a_1 - 1} + 2\sqrt{a_2 - 4} + 3\sqrt{a_3 - 9} = (\dfrac{a_1 + a_2 + a_3}{2})}\\\\\mathsf{\implies \sqrt{2 - 1} + 2\sqrt{8 - 4} + 3\sqrt{a_3 - 9} = (\dfrac{2 + 8 + a_3}{2})}\\\\\mathsf{\implies 5 + 3\sqrt{a_3 - 9} = (5 + \dfrac{a_3}{2})}\\\\\mathsf{\implies 3\sqrt{a_3 - 9} = (\dfrac{a_3}{2})}\\\\\mathsf{Squaring\;on\;both\;Sides,\;We\;get:}$

$\mathsf{\implies 36(a_3 - 9) = (a_3)^2}\\\\\mathsf{\implies (a_3)^2 - 36(a_3) + 324}\\\\\mathsf{\implies (a_3 - 18)^2 = 0}\\\\\mathsf{\implies a_3 = 18}$

$\mathsf{Now,\;Consider\;the\;Situation : n = 4}$

$\mathsf{\implies \sqrt{a_1 - 1} + 2\sqrt{a_2 - 4} + 3\sqrt{a_3 - 9} + 4\sqrt{a_4 - 16} = (\dfrac{a_1 + a_2 + a_3 + a_4}{2})}\\\\\mathsf{\implies \sqrt{2 - 1} + 2\sqrt{8 - 4} + 3\sqrt{18 - 9} + 4\sqrt{a_4 - 16} = (\dfrac{2 + 8 + 18 + a_4}{2})}\\\\\mathsf{\implies 14 + 4\sqrt{a_4 - 16} = (14 + \dfrac{a_4}{2})}\\\\\mathsf{\implies 4\sqrt{a_4 - 16} = (\dfrac{a_4}{2})}\\\\\mathsf{Squaring\;on\;both\;Sides,\;We\;get:}$

$\mathsf{\implies 64(a_4 - 16) = (a_4)^2}\\\\\mathsf{\implies (a_4)^2 - 64(a_4) + 1024}\\\\\mathsf{\implies (a_4 - 32)^2 = 0}\\\\\mathsf{\implies a_4 = 32}$

$\mathsf{Now,\;Consider\;the\;Situation : n = 5}$

$\mathsf{\implies \sqrt{a_1 - 1} + 2\sqrt{a_2 - 4} + 3\sqrt{a_3 - 9} + 4\sqrt{a_4 - 16} + 5\sqrt{a_5 - 25} = (\dfrac{a_1 + a_2 + a_3 + a_4 + a_5}{2})}\\\\\mathsf{\implies \sqrt{2 - 1} + 2\sqrt{8 - 4} + 3\sqrt{18 - 9} + 4\sqrt{32 - 16} + 5\sqrt{a_5 - 25} = (\dfrac{2 + 8 + 18 + 32 + a_5}{2})}\\\\\mathsf{\implies 30 + 5\sqrt{a_5 - 25} = (30 + \dfrac{a_5}{2})}\\\\\mathsf{\implies 5\sqrt{a_5 - 25} = (\dfrac{a_5}{2})}\\\\\mathsf{Squaring\;on\;both\;Sides,\;We\;get:}$

$\mathsf{\implies 100(a_5 - 25) = (a_5)^2}\\\\\mathsf{\implies (a_5)^2 - 100(a_5) + 2500}\\\\\mathsf{\implies (a_5 - 50)^2 = 0}\\\\\mathsf{\implies a_5 = 50}$

$\mathsf{\implies a_1 + a_2 + a_3 + a_4 + a_5 = (2 + 8 + 18 + 32 + 50) = 110}$