If Un - n 15 n 22 N -12 N -1 n 33 N 23 N- then - Un - n -1 NA- N -2B- NC- ND- N -1

The correct option is B ∑nn=1N=N(N+1)2∑n2n=1N=N(N+1)(2N+1)6∑n3n=1N=N(N+1)22∆=∑n=1NUn=NN+11261522N+12N+12N+13NN+13N23NOn nbsp;Applying nbsp;C1→C1 C3, nbsp; ...

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Byju's Answer

Standard XII

Mathematics

Summation of Determinant

If Un = n 15 ...

Question

$If U_{n} = ⎛ ⎜ ⎝ \begin{matrix} n & 1 & 5 n^{2} & 2 N + 1 & 2 N + 1 n^{3} & 3 N^{2} & 3 N \end{matrix} ∣ ∣ ∣ ∣ ∣, then N \sum U_{n} = n = 1$
.

N + 2

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0.0

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N + 1

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Solution

The correct option is B 0.0
$N \sum n n = 1 = \frac{N (N + 1)}{2} N \sum n^{2} n = 1 = \frac{N (N + 1) (2 N + 1)}{6} N \sum n^{3} n = 1 = {(\frac{N (N + 1)}{2})}^{2} Δ = \sum_{n = 1}^{N} U_{n} = \frac{N (N + 1)}{12} ⎛ ⎜ ⎝ \begin{matrix} 6 & 1 & 5 2 (2 N + 1) & 2 N + 1 & 2 N + 1 3 N (N + 1) & 3 N^{2} & 3 N \end{matrix} ∣ ∣ ∣ ∣ ∣ On Applying C_{1} \to C_{1} - C_{3}, we get - Δ = ⎛ ⎜ ⎝ \begin{matrix} 1 & 1 & 5 2 N + 1 & 2 N + 1 & 2 N + 1 3 N^{2} & 3 N^{2} & 3 N \end{matrix} ∣ ∣ ∣ ∣ ∣ As C_{1} and C_{2} are identical Hence, Δ = 0$

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Similar questions

If U_{n} = ⎛ ⎜ ⎝ \begin{matrix} n & 1 & 5 n^{2} & 2 N + 1 & 2 N + 1 n^{3} & 3 N^{2} & 3 N \end{matrix} ∣ ∣ ∣ ∣ ∣, then N \sum U_{n} = n = 1

Q. Evaluate

N \sum n = 1 U_{n}

U_{n} =

∣ ∣ ∣ ∣ ∣ \begin{matrix} n & 1 & 5 n^{2} & 2 N + 1 & 2 N + 1 n^{3} & 3 N^{2} & 3 N \end{matrix} ∣ ∣ ∣ ∣ ∣

Q.
If

U_{n} = sin (n θ) {sec}^{n} θ, V_{n} = cos (n θ) {sec}^{n} θ

for

n = 0, 1, 2

then

V_{n} - V_{n - 1} + U_{n - 1} tan θ =

Q. If

U_{n} = \int_{0}^{1} x^{n} (2 - x)^{n} d x

and

V_{n} = \int_{0}^{1} x^{n} (1 - x)^{n} d x, n \in N

and

\frac{U_{n}}{V_{n}} = 1024,

then the value of

n

is equal to

Q. If

U_{n} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & k & k 2 n & k^{2} + k + 1 & k^{2} + k 2 n - 1 & k^{2} & k^{2} + k + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

and

\sum_{n = 1}^{k} U_{n} = 72

then

k =

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If Un=⎛⎜⎝n15n22N+12N+1n33N23N∣∣ ∣ ∣∣,then N∑Un=n=1 .

The correct option is B 0.0N∑nn=1=N(N+1)2N∑n2n=1=N(N+1)(2N+1)6N∑n3n=1=(N(N+1)2)2Δ=∑Nn=1Un=N(N+1)12⎛⎜⎝6152(2N+1)2N+12N+13N(N+1)3N23N∣∣ ∣ ∣∣On Applying C1→C1−C3, we get−Δ=⎛⎜⎝1152N+12N+12N+13N23N23N∣∣ ∣ ∣∣As C1 and C2 are identicalHence, Δ=0

$If U_{n} = ⎛ ⎜ ⎝ \begin{matrix} n & 1 & 5 n^{2} & 2 N + 1 & 2 N + 1 n^{3} & 3 N^{2} & 3 N \end{matrix} ∣ ∣ ∣ ∣ ∣, then N \sum U_{n} = n = 1$
.