Final answer:
a. The sum of the series 1 + 3 + 5 + 7 + ... + 999 is 250000. b. The sum of the series 2 + 4 + 8 + 16 + ... + 1024 is 4094. c. The sum of the series n(1+i=3)1 represents a formula for the sum of consecutive terms from 3 to n.
Step-by-step explanation:
a. The given series is an arithmetic progression with a common difference of 2. To find the sum, we can use the arithmetic series formula: Sn = (n/2)(a + l), where Sn is the sum, n is the number of terms, a is the first term, and l is the last term. Substituting the values, we get Sn = (500/2)(1 + 999) = 250(1000) = 250000.
b. The given series is a geometric progression with a common ratio of 2. To find the sum, we can use the geometric series formula: Sn = a(1 - r^n) / (1 - r), where Sn is the sum, a is the first term, r is the common ratio, and n is the number of terms. Substituting the values, we get Sn = 2(1 - 2^11) / (1 - 2) = 2(1 - 2048) / -1 = -4094 / -1 = 4094.
c. The given sum is a mathematical expression without specific values. It represents a formula for the sum of consecutive terms from 3 to n. To find the sum, we can use the arithmetic series formula with variable values.
d. The given sum is a mathematical expression without specific values. It represents a formula for the sum of consecutive terms from 3 to n. To find the sum, we can use the arithmetic series formula with variable values.
e. The given sum represents the sum of products of i and (i+1) for values of i from 0 to (n-1). To find the sum, we can use the summation formula: Sn = ∑(i(i+1)), where Sn is the sum, and ∑ represents the summation notation. Simplifying the expression, we get Sn = ∑(i^2 + i), which can be further evaluated using the formulas for the sum of squares and the sum of arithmetic series.
f. The given summation represents the sum of the product of 3 and j where j varies from 1 to n. To find the sum, we can use the formula for the sum of arithmetic progression: Sn = (n/2)(a + l), where Sn is the sum, n is the number of terms, a is the first term, and l is the last term. Substituting the values, we get Sn = (n/2)(3 + 3n). This can be further simplified by expanding the expression and combining like terms.
g. The given double summation represents the sum of products of i and j for all possible values of i from 1 to n and j from 1 to n. To find the sum, we can use nested summations and the distributive property to expand the expression. The final expression can be further simplified by arranging the terms and combining like terms.
h. The given summation represents the sum of the reciprocal of i(i-1) for values of i from 1 to n. To find the sum, we can use the formula for the sum of reciprocal of squares: Sn = ∑(1 / (i(i-1))), where Sn is the sum, and ∑ represents the summation notation. Simplifying the expression, we get Sn = ∑(1 / (i-1) - 1 / i), which can be evaluated using the telescoping series method.