Question

Let $\sf{T_r}$ be the $\sf{r^{th}}$ term of a sequence, for $\sf{r=1,2,3,4\cdots}$. If $\sf{3\:T_{r+1}=T_r}$ and $\sf{\dfrac{1}{243}}$, then find the value of $\sf{\left(i\right)\:\Sigma_{r=1}^5\:\dfrac{1}{T_{r+1}}} \\\\\\ \sf{\left(ii\right)\:\Sigma_{r=1}^{\infty}\left(T_r\:.\:T_{r+1}\right)}$

Answer 1

Correct question :-

$\sf{ \implies \: T _r = rth \:term \: }$

$\sf{ \implies \: 3T_{r + 1} = T _r }$

$\sf{ \implies \: T _7 \: = \dfrac{1}{243} }$

To Find :-

$\sf{\implies \sum_{r = 1}^{5} \: \dfrac{1}{T _{r + 1} }}$

$\sf { \implies \: \Sigma_{r=1}^{\infty}\left(T_r\:.\:T_{r+1}\right)}$

Solution :-

Part 2nd :-

$\sf{ \implies \: 3T_{r + 1} = T _r }$

$\sf{ \implies \: \dfrac{T_{r + 1}}{T_r} = \dfrac{1}{3} }$

Through this we find out the common ratio of this Sequence ie 1/3

Let's assume that $\sf{T_1 \: = y }$

$\sf{\implies T_n \: = a {r}^{n - 1} }$

$\sf{\implies T_2 = \frac{1}{3} \times y}$

$\sf{\implies T_3 = y \times \dfrac{1}{{3}^{2}} }$

$\sf{\implies T_4 = y \times \dfrac{1}{{3}^{3}} }$

Similarly we'll find the value of 7th term

$\sf{\implies T_7 = y \times \dfrac{1}{{3}^{6}} }$

$\sf{\implies T_7 = \dfrac{y}{{3}^{6}} = \dfrac{1}{243} }$

$\sf{\implies T_7 = \dfrac{y}{{3}^{6}} = \dfrac{1}{{3}^{5} } }$

$\sf{ \implies \: \dfrac{1}{ \cancel{ {3}^{5} }} = \dfrac{y}{ \cancel{{3}^{6} }} }$

$\underline{\underline{\sf{\implies y = 3 }}}$

$\sf{ \implies \:T _r. \times T _{r + 1} = \dfrac{y}{ {3}^{r - 1} } \times \dfrac{y}{ {3}^{r} } }$

$\sf{ \implies \: \dfrac{ {y}^{2} }{ {3}^{2r - 1} } = \dfrac{ {3}^{2} }{ {3}^{2r - 1} } }$

$\sf { \implies \: \sum_{r = 1}^{ \infty} = T_r.T_{r + 1}}$

$\sf{ \implies \:3 + \frac{1}{3} + \frac{1}{ {3}^{3} } + \frac{1}{ {3}^{5} } ....... \infty}$

Sum of infinite G.P -

$\sf{ \implies \: S _{ \infty} \: = \dfrac{a}{1 - r} }$

$\sf{ \implies \: S_{ \infty} \: = \dfrac{3}{1 - \dfrac{1}{{3}^{2}}} }$

$\underline{ \underline{ \sf{ \red{ \implies \: \dfrac{27}{8} }}}}$

Part 1st :-

$\sf{\implies \sum_{r = 1}^{5} \: \dfrac{1}{T _{r + 1} }}$

$\sf{\implies \dfrac{1}{T_{1+1}} + \dfrac{1}{T_{2+1}} + \dfrac{1}{T_{3+1} }+\dfrac{1}{T_{4+1} }+\dfrac{1}{T_{5+1}}}$

$\sf{\implies T_1 = 3}$

$\sf{\implies T_2 = \frac{1}{3} \times 3 = 1 }$

$\sf{\implies T_3 = \frac{1}{{3}^{2}} \times 3 = \dfrac{1}{3} }$

$\sf{\implies T_4 = \frac{1}{{3}^{3}} \times 3 = \dfrac{1}{{3}^{2} }}$

$\sf{\implies T_5 = \frac{1}{{3}^{4}} \times 3 = \dfrac{1}{{3}^{3} }}$

$\sf{\implies T_6 = \frac{1}{{3}^{5}} \times 3 = \dfrac{1}{{3}^{4} }}$

$\sf{\implies \dfrac{1}{1 } + \dfrac{1}{\dfrac{1}{3}} + \dfrac{1}{\dfrac{1}{{3}^{2}}} +\dfrac{1}{\dfrac{1}{{3}^{3}}} +\dfrac{1}{\dfrac{1}{{3}^{4}}} }$

$\sf{\implies \dfrac{1}{1} + \dfrac{3}{1} + \dfrac{{3}^{2}}{1} +\dfrac{{3}^{3}}{1} +\dfrac{{3}^{4}}{1}}$

$\sf{\implies \: 1 + 3 + 9 + 27 + 81 }$

$\underline{ \underline{ \sf{ \red{ \implies \: 121 }}}}$

Answer 2

Step-by-step explanation:

Answer:

To integrate the rationalrational function x(x−1)(x−2)(x−3)x(x−1)(x−2)(x−3)

Let integrand x(x−1)(x−2)(x−3)=Ax−1+Bx−2+Cx−3…(i)x(x−1)(x−2)(x−3)=Ax−1+Bx−2+Cx−3…(i) PartialfractionsPartialfractions

Multiplying by L. C.M. =(x−1)(x−2)(x−3) , x=Λ(x−2)(x−3)+B(x−1)(x−3)+C(x−1)(x−2)=A(x2−5x+6)+B(x2−4x+3)+C(x2−3x+2)=Ax2−6Ax+6A+Bx2−4Bx+3B+Cx2−3Cx+2C

Multiplying by L. C.M. =(x−1)(x−2)(x−3) , x=Λ(x−2)(x−3)+B(x−1)(x−3)+C(x−1)(x−2)=A(x2−5x+6)+B(x2−4x+3)+C(x2−3x+2)=Ax2−6Ax+6A+Bx2−4Bx+3B+Cx2−3Cx+2C

Comparing coefficients of x2,xx2,x and constant terms on both aides, we have x2x2 A+B+C=0A+B+C=0 xx ? −5A−4B−3C=1−5A−4B−3C=1 or 5A+4B+3C=−15A+4B+3C=−1

Constants: 6A+3B+2C=06A+3B+2C=0 Let us solve Eqns. (ii),(iii)(ii),(iii) and (iv)(iv) for A,B,C.A,B,C.

Let us first form two Eqns. in two unknowns say A and B.

Eqn. (ii)−3×(ii)−3× Eqn. iiii gives

5A+4B+3C−3A−3B−3C=−15A+4B+3C−3A−3B−3C=−1 or 2A+B=−12A+B=−1 …(v)…(v)

Eqn. (iv)−2×(iv)−2× Eqn. iiii gives

4A+B=04A+B=0

Eqn. (vi)−(vi)− Eqn. 00 gives ToeliminateBToeliminateB

2A=1∴A=122A=1∴A=12

Putting A=12A=12 in (v),1+B=−1⇒B=−2(v),1+B=−1⇒B=−2

Putting A=12A=12 and B=−2B=−2 in (ii)(ii)

12−2+C=0⇒C=−12+2=−1+42=3212−2+C=0⇒C=−12+2=−1+42=32

Putting these values of A, B, C in ii, we have

x(x−1)(x−2)(x−3)=12x−1−2x−2+32x−3

x(x−1)(x−2)(x−3)=12x−1−2x−2+32x−3

∴∫x(x−1)(x−2)(x−3)dx∴∫x(x−1)(x−2)(x−3)dx =12∫1x−1dx−2∫1x−2dx+32∫1x−3dx=12∫1x−1dx−2∫1x−2dx+32∫1x−3dx

=12log|x−1|−2log|x−2|+32log|x−3|+c

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