If the value of -r-210r2-r-1r-1rr-1r-2 is k40- then k is equal to

∑_r=2^10r^2+r+1(r 1)(r)(r+1)(r+2) =∑_r=2^10r(r 1)(r+1) r+1r(r+2) =21·3 32·4+32·4 43·5+⋯+109·11 1110·12 =21·3 1110·12 =2340=k40 ∴ k=23 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Sum of N Terms of an AP

If the value ...

Question

If the value of $10 \sum r = 2 \frac{r^{2} + r + 1}{(r - 1) (r) (r + 1) (r + 2)}$ is $\frac{k}{40},$ then $k$ is equal to

29

No worries! We‘ve got your back. Try BYJU‘S free classes today!

27

No worries! We‘ve got your back. Try BYJU‘S free classes today!

23

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

21

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $23$
$10 \sum r = 2 \frac{r^{2} + r + 1}{(r - 1) (r) (r + 1) (r + 2)}$
$= 10 \sum r = 2 \frac{r}{(r - 1) (r + 1)} - \frac{r + 1}{r (r + 2)}$
$= \frac{2}{1 \cdot 3} - \frac{3}{2 \cdot 4} + \frac{3}{2 \cdot 4} - \frac{4}{3 \cdot 5} + \dots + \frac{10}{9 \cdot 11} - \frac{11}{10 \cdot 12}$
$= \frac{2}{1 \cdot 3} - \frac{11}{10 \cdot 12}$
$= \frac{23}{40} = \frac{k}{40}$
$∴ k = 23$

Suggest Corrections

Similar questions

Which of the following best describes the terms of the series? $\frac{1}{2}$ $^{n} C_{1}$ - $\frac{2}{3}$ $^{n} C_{2}$ + $\frac{3}{4}$ $^{n} C_{3}$ ................... $(- 1)^{n + 1}$ $\frac{1}{n + 1}$ $^{n} C_{n}$

Q. The value of

\sum_{r = 1}^{n} {(- 1)}^{r + 1} \frac{^{n} C_{r}}{r + 1}

is equal to

Q. If

r = n \sum r = 1 \frac{r^{4} + r^{2} + 1}{r^{4} + r} = \frac{675}{26},

then

n

is equal to

Q. The value of

100 \sum r = 2 \frac{3^{r} (2 - 2 r)}{(r + 1) (r + 2)}

is equal to?

Q. The value of

lim n \to \infty r = 4 n \sum r = 1 \frac{\sqrt{n}}{\sqrt{r} {(3 \sqrt{r} + \sqrt{n})}^{2}}

is equal to

Join BYJU'S Learning Program

If the value of 10∑r=2r2+r+1(r−1)(r)(r+1)(r+2) is k40, then k is equal to

The correct option is C 2310∑r=2r2+r+1(r−1)(r)(r+1)(r+2) =10∑r=2r(r−1)(r+1)−r+1r(r+2) =21⋅3−32⋅4+32⋅4−43⋅5+⋯+109⋅11−1110⋅12 =21⋅3−1110⋅12 =2340=k40 ∴k=23

If the value of $10 \sum r = 2 \frac{r^{2} + r + 1}{(r - 1) (r) (r + 1) (r + 2)}$ is $\frac{k}{40},$ then $k$ is equal to