- r -1 n T r - n -8 n -1 n -2 n -3- and - r -1 n 1- T r - n 2-3 n -4 - k -1 p k- then p is equal toA- n-1B- nC- 2 nD- n-1

The correct option is A n+1∵ T_n =S_n S_n 1 =∑_r=1^n T_r ∑_r=1^n 1T_r = n(n+1)(n+2)(n+3)8 (n 1)n(n+1)(n+2)8 =n(n+1)(n+2)2 ⇒1T_n =2n(n+1)(n+2) = (n+2) n ...

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Byju's Answer

Standard XII

Mathematics

Method of Difference

∑ r =1 n T r ...

Question

If $n \sum r = 1 T_{r} = \frac{n}{8} (n + 1) (n + 2) (n + 3)$ , and $n \sum r = 1 \frac{1}{T_{r}} = \frac{n^{2} + 3 n}{4 p \sum k = 1 k}$ , then $p$ is equal to

n + 1

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n

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n - 1

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2 n

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Solution

The correct option is A $n + 1$
$∵ T_{n} = S_{n} - S_{n - 1}$
$= n \sum r = 1 T_{r} - n - 1 \sum r = 1 T_{r} = \frac{n (n + 1) (n + 2) (n + 3)}{8} - \frac{(n - 1) n (n + 1) (n + 2)}{8}$
$= \frac{n (n + 1) (n + 2)}{2}$
$\Rightarrow \frac{1}{T_{n}} = \frac{2}{n (n + 1) (n + 2)} = \frac{(n + 2) - n}{n (n + 1) (n + 2)} = \frac{1}{n (n + 1)} - \frac{1}{(n + 1) (n + 2)} \dots (i)$
Let $V_{n} = \frac{1}{n (n + 1)}$
$∴ \frac{1}{T_{r}} = V_{r} - V_{r + 1} n \sum r = 1 \frac{1}{T_{r}} = (V_{1} - V_{n + 1}) \Rightarrow n \sum r = 1 \frac{1}{T_{r}} = \frac{n^{2} + 3 n}{2 (n + 1) (n + 2)}$
and $4 p \sum k = 1 k = 2 p (p + 1) = 2 (n + 1) (n + 2)$
$∴ p = n + 1$

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Similar questions

Q. If

n \sum r = 1 T_{r} = \frac{n}{8} (n + 1) (n + 2) (n + 3)

and

n \sum r = 1 \frac{1}{T_{r}} = \frac{n^{2} + 3 n}{4 p \sum k = 1 k}

, then

p

is equal to

Q. If

n \sum r = 1 T_{r} = \frac{n}{8} (n + 1) (n + 2) (n + 3)

and

n \sum r = 1 \frac{1}{T_{r}} = \frac{n^{2} + 3 n}{4 p \sum k = 1 k}

, then

p

is equal to

Q. If

\sum_{r = 1}^{n} T_{r} = \frac{n}{8} (n + 1) (n + 2) (n + 3)

then find

\sum_{r = 1}^{n} \frac{1}{T_{r}}

Q. If

n \sum r = 1 T_{r} = \frac{n}{8} (n + 1) (n + 2) (n + 3)

, then find

n \sum r = 1 \frac{1}{T_{r}}

Q. If

n \sum r = 1 t_{r} = n \sum k = 1 k \sum j = 1 j \sum i = 1 (2)

, then

n \sum r = 1 {\frac{1}{t}}_{r}

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If n∑r=1Tr=n8(n+1)(n+2)(n+3), and n∑r=11Tr=n2+3n4p∑k=1k, then p is equal to

If $n \sum r = 1 T_{r} = \frac{n}{8} (n + 1) (n + 2) (n + 3)$ , and $n \sum r = 1 \frac{1}{T_{r}} = \frac{n^{2} + 3 n}{4 p \sum k = 1 k}$ , then $p$ is equal to