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Telescoping Series with Triangular, Odd and Cube Sums to 10 Terms | JEE

JEE Maths question with a full step-by-step solution.

Question

1 + \frac{(1+2)(1+3)}{2^{2}\left(1^{3}+2^{3}\right)} + \frac{(1+2+3)(1+3+5)}{3^{2}\left(1^{3}+2^{3}+3^{3}\right)} + \frac{(1+2+3+4)(1+3+5+7)}{4^{2}\left(1^{3}+2^{3}+3^{3}+4^{3}\right)} + \cdots

up to

10

terms is equal to

\dfrac{a}{b}

, where

a

and

b

are relatively prime, then

a - b

is equal to

Solution

Answer: 9

Step 1: Write the general term. Reading off the pattern, the

r

-th term is

T_r = \frac{(1+2+\cdots+r)\left(1+3+\cdots+(2r-1)\right)}{r^{2}\left(1^{3}+2^{3}+\cdots+r^{3}\right)} .

Step 2: Replace each block by its standard formula.

1+2+\cdots+r = \frac{r(r+1)}{2}, \qquad 1+3+\cdots+(2r-1) = r^{2}, \qquad 1^{3}+\cdots+r^{3} = \frac{r^{2}(r+1)^{2}}{4} .

Step 3: Substitute.

T_r = \frac{\dfrac{r(r+1)}{2}\cdot r^{2}}{r^{2}\cdot \dfrac{r^{2}(r+1)^{2}}{4}} .

Step 4: Cancel. The

r^{2}

in the numerator cancels the first

r^{2}

below, and multiplying by the reciprocal of the remaining fraction,

T_r = \frac{r(r+1)}{2}\cdot\frac{4}{r^{2}(r+1)^{2}} = \frac{2}{r(r+1)} .

Step 5: Check against the printed first term:

T_1 = \dfrac{2}{1\times 2} = 1

, and

T_2 = \dfrac{2}{2\times 3} = \dfrac13

, which matches

\dfrac{3 \times 4}{4 \times 9} = \dfrac13

. Step 6: Split into a telescoping difference.

T_r = 2\left(\frac{1}{r} - \frac{1}{r+1}\right).

Step 7: Add the first ten terms; all the middle pieces cancel.

\sum_{r=1}^{10} T_r = 2\left(\frac11 - \frac{1}{11}\right) = 2 \times \frac{10}{11} = \frac{20}{11} .

Step 8: Here

a = 20

and

b = 11

, which are relatively prime, so

a - b = 20 - 11 = 9 .

Answer:

9

(i.e.

9.00

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