Sum of series - r -1 n r 2-1 r - isA- n -2 -1B- n n -1 -C- n n -1 -1D- n -1 -

The correct option is B n(n+1)!T_r = (r^2 + 1 + r r) r! =r(r+1)r! (r 1)r! = r(r + 1)! (r 1)r! ⇒ T_1=1× 2! 0 T_2=2× 3! 1× 2! T_3=3× 4! 2× 3! T_4=4× 5! 3× ...

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Byju's Answer

Standard XII

Mathematics

Geometric Progression

Sum of series...

Question

Sum of series $\sum_{r = 1}^{n} (r^{2} + 1) r!$ is

(n + 1)!

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(n + 2)! - 1

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n (n + 1)!

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n (n + 1)! - 1

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Solution

The correct option is C $n (n + 1)!$
$T_{r} = (r^{2} + 1 + r - r) r!$
$= r (r + 1) r! - (r - 1) r!$
$= r (r + 1)! - (r - 1) r!$
$\Rightarrow T_{1} = 1 \times 2! - 0$
$T_{2} = 2 \times 3! - 1 \times 2!$
$T_{3} = 3 \times 4! - 2 \times 3!$
$T_{4} = 4 \times 5! - 3 \times 4!$
.
.
.
$T_{n - 1} = (n - 1) \times n! - (n - 2) \times (n - 1)!$
$T_{n} = n \times (n + 1)! - (n - 1) \times n!$
Now, $S_{n} = T_{1} + T_{2} + T_{3} + . . . + T_{n - 1} + T_{n}$
$= n (n + 1)!$

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Sum of series ∑nr=1(r2+1)r! is

The correct option is C n(n+1)!Tr=(r2+1+r−r)r! =r(r+1)r!−(r−1)r! =r(r+1)!−(r−1)r! ⇒T1=1×2!−0 T2=2×3!−1×2! T3=3×4!−2×3! T4=4×5!−3×4! . . . Tn−1=(n−1)×n!−(n−2)×(n−1)! Tn=n×(n+1)!−(n−1)×n! Now, Sn=T1+T2+T3+...+Tn−1+Tn =n(n+1)!

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