Find general term of series loge(1+3x+2x2)=3x−25x2+39x3−417x4+..

loge(1+3x+2x2)=3x−25x2+39x3−417x4+..Now,1+3x+2x2=2x2+2x+x+1 =2x(x+1)+1(x+1) =(2x+1)(x+1)loge(1+3x+2x2)=loge[(1+2x)(1+x)] =loge(1+2x)+loge(1+x)Now,loge(1+2x)=2x−2(2x)2+3(2x)3−4(2x)4+.... --- ( 1 )loge(1+x)=x−2(x)2+3(x)3−4(x)4−.... ---( 2 )Adding ( 1 ) from ( 2 ) we get,loge(1+2x)+loge(1+x)=[2x−2(2x)2+3(2x)3−4(2x)4+....]+[x−2(x)2+3(x)3−4(x)4−....]loge(1+2x)+loge(1+x)=3x−(24x2+x2)+(38x3+x3)−(416x4+x4)+...loge(1+2x)+loge(1+x)=3x−25x2+39x3−417x4+... General term of series =(−1)n−1(n2xn+xn)

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Find general term of series _e(1+3x+2x^2)=3x-{5x^2}{2}+{9x^3

Question

Find general term of series
$lo g_{e} (1 + 3 x + 2 x^{2}) = 3 x - \frac{5 x ^{2}}{2} + \frac{9 x ^{3}}{3} - \frac{17 x ^{4}}{4} + ..$

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Find general term of series

lo g_{e} (1 + 3 x + 2 x^{2}) = 3 x - \frac{5 x ^{2}}{2} + \frac{9 x ^{3}}{3} - \frac{17 x ^{4}}{4} + ..

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Text solutionVerified

lo g_{e} (1 + 3 x + 2 x^{2}) = 3 x - \frac{5 x ^{2}}{2} + \frac{9 x ^{3}}{3} - \frac{17 x ^{4}}{4} + ..

Now,

1 + 3 x + 2 x^{2} = 2 x^{2} + 2 x + x + 1

= 2 x (x + 1) + 1 (x + 1)

= (2 x + 1) (x + 1)

lo g_{e} (1 + 3 x + 2 x^{2}) = lo g_{e} [(1 + 2 x) (1 + x)]

= lo g_{e} (1 + 2 x) + lo g_{e} (1 + x)

Now,

lo g_{e} (1 + 2 x) = 2 x - \frac{( 2 x ) ^{2}}{2} + \frac{( 2 x ) ^{3}}{3} - \frac{( 2 x ) ^{4}}{4} + ....

--- ( 1 )

lo g_{e} (1 + x) = x - \frac{( x ) ^{2}}{2} + \frac{( x ) ^{3}}{3} - \frac{( x ) ^{4}}{4} - ....

---( 2 )

Adding ( 1 ) from ( 2 ) we get,

lo g_{e} (1 + 2 x) + lo g_{e} (1 + x) = [2 x - \frac{( 2 x ) ^{2}}{2} + \frac{( 2 x ) ^{3}}{3} - \frac{( 2 x ) ^{4}}{4} + ....] + [x - \frac{( x ) ^{2}}{2} + \frac{( x ) ^{3}}{3} - \frac{( x ) ^{4}}{4} - ....]

lo g_{e} (1 + 2 x) + lo g_{e} (1 + x) = 3 x - (\frac{4 x ^{2} + x ^{2}}{2}) + (\frac{8 x ^{3} + x ^{3}}{3}) - (\frac{16 x ^{4} + x ^{4}}{4}) + ...

lo g_{e} (1 + 2 x) + lo g_{e} (1 + x) = 3 x - \frac{5 x ^{2}}{2} + \frac{9 x ^{3}}{3} - \frac{17 x ^{4}}{4} + ...

General term of series

= (- 1)^{n - 1} (\frac{2 x ^{n} + x ^{n}}{n})

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Question Text	Find general term of series $lo g_{e} (1 + 3 x + 2 x^{2}) = 3 x - \frac{5 x ^{2}}{2} + \frac{9 x ^{3}}{3} - \frac{17 x ^{4}}{4} + ..$
Answer Type	Text solution:1
Upvotes	150

Find general term of series loge​(1+3x+2x2)=3x−25x2​+39x3​−417x4​+..

Text solutionVerified

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Find general term of series
$lo g_{e} (1 + 3 x + 2 x^{2}) = 3 x - \frac{5 x ^{2}}{2} + \frac{9 x ^{3}}{3} - \frac{17 x ^{4}}{4} + ..$