Question

pls answer this question​

Answer 1

Answer: 25.

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Answer 2

Answer:

$\sum_{r=1}^{50} \cfrac{r^2}{r^2 + (11-r)^2} = 2.86 \text{ approx.}$

Step-by-step explanation:

$\sum_{r=1}^{50} \cfrac{r^2}{r^2 + (11-r)^2} = \sum_{r=1}^{50} \cfrac{r^2}{r^2 + 121 +r^2 -22r}$

$=\sum_{r=1}^{50}\cfrac{r^2}{2r^2 - 22r + 121}$

$=\sum_{r=1}^{50} \cfrac{1^2+2^2+3^2+...+50^2}{2(1^2+2^2+3^2+...+50^2)-22(1+2+3+...+50)+121}$     ....(1)

We know that sum of square of first n natural numbers is given by

$\sum \cfrac{n(n+1)(2n+1)}{6}$

and sum of first n natural numbers is given by

$\sum \cfrac{n(n+1)}{2}$

So, Eqn. (1) becomes,

$\sum_{r=1}^{50} \cfrac{r^2}{r^2 + (11-r)^2} = \cfrac{\cfrac{50(50+1)(2\times50+1}{6}}{2\left[ \cfrac{50(50+1)(2\times50+1}{6} \right]-22\left[\cfrac{50(50+1)}{2} \right] + 121}$

$=\cfrac{\cfrac{50\times51\times101}{6}}{2\left(\cfrac{50\times51\times101}{6}\right) - 22(25)(51) + 121}$

$=\cfrac{\cfrac{25\times\51\times101}{3}}{\cfrac{50\times51\times101}{3}-22\times25\times51 + 121}$

$=\cfrac{25\times17\times101}{25\times17\times101-22\times25\times51+121}$

$=\cfrac{42925}{42925-28050+121}$

$=\cfrac{42925}{14996} = 2.86 \text{ approx.}$

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