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Q.If t=45+4.65.10+4.6.85.10.15+…∞t=\dfrac{4}{5}+\dfrac{4.6}{5.10}+\dfrac{4.6.8}{5.10.15}+\ldots\infty, then prove that 9t=169t=16.

Telangana TsbieTelangana Board of Intermediate Education 2020Subjective· 7mImportance★★★★★
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Recognize the general term of the series as (k+1)(25)k(k+1)\left(\dfrac25\right)^k and sum it using the standard series ∑kxk\sum kx^k and ∑xk\sum x^k.

The series is

t=45+4⋅65⋅10+4⋅6⋅85⋅10⋅15+⋯∞t = \frac{4}{5}+\frac{4\cdot6}{5\cdot10}+\frac{4\cdot6\cdot8}{5\cdot10\cdot15}+\cdots\infty

Find the general (kthk^{th}) term, k=1,2,3,…k=1,2,3,\ldots. The numerator of the kthk^{th} term is the product of kk factors starting at 44, increasing by 22: 4⋅6⋯(2k+2)=2k (2⋅3⋯(k+1))=2k(k+1)!4\cdot6\cdots(2k+2) = 2^k\,(2\cdot3\cdots(k+1)) = 2^k(k+1)!.

The denominator is the product of kk multiples of 55: 5⋅10⋯5k=5k k!5\cdot10\cdots5k = 5^k\,k!.

So the kthk^{th} term is

Tk=2k(k+1)!5k k!=(k+1)(25)kT_k = \frac{2^k(k+1)!}{5^k\,k!} = (k+1)\left(\frac25\right)^k

(Check k=1k=1: T1=2⋅25=45T_1=2\cdot\frac25=\frac45 ✓. Check k=2k=2: T2=3(25)2=3⋅425=1225=2450T_2=3\left(\frac25\right)^2=3\cdot\frac{4}{25}=\frac{12}{25}=\frac{24}{50} ✓.)

So t=∑k=1∞(k+1)xkt = \displaystyle\sum_{k=1}^{\infty}(k+1)x^k with x=25x=\dfrac25 (and ∣x∣<1|x|<1, so the series converges).

Split it:

t=∑k=1∞kxk+∑k=1∞xkt = \sum_{k=1}^\infty kx^k + \sum_{k=1}^\infty x^k

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