. Below series chooses. This code segment find thevalue of the result of followingseries for five terms. *YX = Val (Text1.Text)N=1: K=1: S=0: C=11-.X2 3X6 5X1022 62 10²+S=S+K*TK=-K: C=C+1O*****.2-3-.....GoTo 2010 Text2.Text = Str(S)O1-20 T = (x^(2*N))/ ((2 *N)^2) 2-N= N +23-1f C = 6 Then Go To 101-20 T = (2*N) * (X^(2*N)) / ((2 * N)^2) 2-N = N + 23-lf C = 6 Then GoTo101-20 T = (N) * (X^(2* N)) / ((2 * N) ^2) 2-N = N + 33-lf C = 6 Then GoTo 101-20 T = (N) * (X^(2 *N)) / ((2 * N) ^2) 2-N = N + 23-1f C = 5 Then GoTo 101-20 T = (N) *(X^(2*N)) / ((2 *N) ^2) 2-N = N +23-lf C = 41= 4 Then GoTo 101-20 T = (N) * (X^(2 * N)) / ((2 * N) ^2) 2-N = N +23-lf C = 6 Then GoTo 201-20 T = (N) * (X^ (2 * N)) / ((2 * N) ^2) 2-N = N +23-1f C = 67= 6 Then GoTo 10

From the question, the variable C maintains the count of no.of terms in it. We need the result for 5…

Answered: low series chooses.

Engineering

Computer Science

low series chooses.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Concept explainers

LUP Decomposition

LU decomposition stands for Lower-Upper decomposition. LU decomposition is the factorization of the lower triangular matrix and an upper triangular matrix. It is very useful in linear algebra and numerical analysis. A permutation matrix is usually included in the product. Gaussian elimination in a matrix form is known as LU decomposition, When inverting a matrix or evaluating its determinant, this is a crucial step. Tadeusz Banachiewicz, a Polish mathematician, first proposed the LU decomposition in 1938.

Question

. Below series chooses.

This code segment find the
value of the result of following
series for five terms. *
Y
X = Val (Text1.Text)
N=1: K=1: S=0: C=1
1-.
X2 3X6 5X10
22 62 10²
+
S=S+K*T
K=-K: C=C+1
O
*****.
2-
3-.....
GoTo 20
10 Text2.Text = Str(S)
O
1-20 T = (x^(2*N))/ ((2 *N)^2) 2-N
= N +23-1f C = 6 Then Go To 10
1-20 T = (2*N) * (X^(2*N)) / ((2 * N)
^2) 2-N = N + 23-lf C = 6 Then GoTo
10
1-20 T = (N) * (X^(2* N)) / ((2 * N) ^
2) 2-N = N + 33-lf C = 6 Then GoTo 10
1-20 T = (N) * (X^(2 *N)) / ((2 * N) ^
2) 2-N = N + 23-1f C = 5 Then GoTo 10
1-20 T = (N) *(X^(2*N)) / ((2 *N) ^
2) 2-N = N +23-lf C = 41
= 4 Then GoTo 10
1-20 T = (N) * (X^(2 * N)) / ((2 * N) ^
2) 2-N = N +23-lf C = 6 Then GoTo 20
1-20 T = (N) * (X^ (2 * N)) / ((2 * N) ^
2) 2-N = N +23-1f C = 67
= 6 Then GoTo 10

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