Free math worksheets

- The lower and upper bounds tell you which integer values of \(k\) to use. - Evaluate the summand separately for each allowed index before adding.

Substitute \(k=1,2,3,4,5\) into \(3k-2\): \(\sum_{k=1}^{5}(3k-2)=1+4+7+10+13=35\).

\(1+4+7+10+13=35\).

- Use the constant difference between consecutive terms to determine the coefficient of \(k\). - Check the proposed summand at the required starting index \(k=1\). - Match the number of displayed terms to the upper bound.

The terms increase by \(4\), so the coefficient of \(k\) is \(4\). To make the first term equal to \(5\) when \(k=1\), use \(4k+1\). There are \(6\) terms, so the sum is \(\sum_{k=1}^{6}(4k+1)\).

\(\sum_{k=1}^{6}(4k+1)\).

- Read both the lower and upper bounds of the sigma notation. - Write the allowed integer index values in order before counting them. - The upper bound is an ending index, not automatically the number of terms.

1. The index starts at \(4\) and increases by \(1\) through \(8\). 2. The index values are \(4,5,6,7,8\). 3. Therefore the sigma expression contains \(5\) terms, not \(8\).

The index values are \(4,5,6,7,8\), so the sum contains \(5\) terms.

- The matching bounds allow the two sums to be combined term by term. - Keep track of the factor outside the second sigma expression when combining summands. - After simplifying to one linear summand, use the sum of the first six positive integers.

1. The two sums have the same bounds, so combine their summands while keeping the factor \(2\): \(\sum_{k=1}^{6}[(4k-3)+2(k+5)]\). 2. Simplify the summand: \((4k-3)+2(k+5)=6k+7\). Thus the expression is \(\sum_{k=1}^{6}(6k+7)\). 3. Evaluate: \(6\sum_{k=1}^{6}k+\sum_{k=1}^{6}7=6\cdot21+7\cdot6=168\).

\(\sum_{k=1}^{6}(6k+7)=168\).

- Matching indices and bounds allow term-by-term combination. - Apply the subtraction sign to every term in the second summand before collecting like terms.

Because the sums have the same index and bounds, subtract their summands: \(\sum_{k=1}^{7}[(2k^2+1)-(k^2-4k+3)]\). Simplifying inside the brackets gives \(2k^2+1-k^2+4k-3=k^2+4k-2\). Therefore the combined expression is \(\sum_{k=1}^{7}(k^2+4k-2)\).

\(\sum_{k=1}^{7}(k^2+4k-2)\).

- Choose a relationship between \(j\) and \(k\) that moves the starting index from \(0\) to \(1\). - Change both bounds using the same index relationship. - Substitute the old index in the summand and check the first and last generated terms.

Let \(j=k+1\), so \(k=j-1\). When \(k=0\), \(j=1\); when \(k=4\), \(j=5\). Replace \(k\) in the summand: \(2k+1=2(j-1)+1=2j-1\). Therefore \(\sum_{k=0}^{4}(2k+1)=\sum_{j=1}^{5}(2j-1)\). Both forms represent \(1+3+5+7+9\).

\(\sum_{j=1}^{5}(2j-1)\).

- Compare the lower bounds of the two sums and identify exactly which indices occur only in the larger one. - Evaluate the summand at the indices that must be removed. - After the sigma manipulation is correct, use a familiar formula or arithmetic-series reasoning to evaluate the larger sum.

1. The larger sum includes the terms for \(k=1\) and \(k=2\), which are not in the original sum. 2. Therefore \(\sum_{k=3}^{12}(2k-1)=\sum_{k=1}^{12}(2k-1)-1-3\). 3. The sum of the first \(12\) odd positive integers is \(12^2=144\). 4. Hence the original sum is \(144-1-3=140\).

\(\sum_{k=3}^{12}(2k-1)=\sum_{k=1}^{12}(2k-1)-1-3=140\).

- A correct split must include every original integer index exactly once. - When two ranges are adjacent, the second range begins one index after the first range ends. - The dummy index letter does not affect whether adjacent ranges can be merged; the summand and coverage of indices do.

a) The indices \(2\) through \(20\) are split between \(7\) and \(8\), so \(\sum_{k=2}^{20}f(k)=\sum_{k=2}^{7}f(k)+\sum_{k=8}^{20}f(k)\). b) The two ranges are adjacent and together contain every integer index from \(4\) through \(15\) exactly once. Therefore \(\sum_{j=4}^{m}g(j)+\sum_{j=m+1}^{15}g(j)=\sum_{j=4}^{15}g(j)\).

a) \(\sum_{k=2}^{7}f(k)+\sum_{k=8}^{20}f(k)\) b) \(\sum_{j=4}^{15}g(j)\)

- Separate the pattern into the magnitude of each term and the sign of each term. - Look for a power of \(-1\) whose value is positive when \(k\) is odd and negative when \(k\) is even. - Test your proposed summand at the first few index values before finalizing the sigma notation.

1. The magnitude of the \(k\)th term is \(k^2\). 2. The sign is positive for odd \(k\) and negative for even \(k\), which is represented by \((-1)^{k+1}\). 3. Therefore the sum is \(\sum_{k=1}^{9}(-1)^{k+1}k^2\). 4. Its first four summands are \(1^2-2^2+3^2-4^2\), matching the required sign pattern.

\(\sum_{k=1}^{9}(-1)^{k+1}k^2\). Its first four summands are \(1^2-2^2+3^2-4^2\).

- Solve the stated index substitution for \(k\) before changing the summand. - Apply the substitution separately to the lower bound, upper bound, and summand. - Simplify only after the reindexed expression is structurally correct.

a) Since \(j=k-3\), we have \(k=j+3\). The old lower bound \(k=3\) becomes \(j=0\), and the old upper bound \(k=8\) becomes \(j=5\). Thus \(\sum_{k=3}^{8}(2k^2-k)=\sum_{j=0}^{5}[2(j+3)^2-(j+3)]\). b) Simplify: \(2(j+3)^2-(j+3)=2j^2+11j+15\). Therefore the sum is \(\sum_{j=0}^{5}(2j^2+11j+15)=365\).

a) \(\sum_{j=0}^{5}[2(j+3)^2-(j+3)]\). b) \(\sum_{j=0}^{5}(2j^2+11j+15)=365\).

- Solve the substitution \(j=k-1\) for the old index \(k\). - Apply the index change to the summand as well as to the bounds. - Compare the first generated term in each expression before evaluating the whole sum.

a) If \(j=k-1\), then \(k=j+1\). The student changed the bounds but failed to replace \(k\) correctly in the summand. b) Since \(k+2=(j+1)+2=j+3\), the correct reindexing is \(\sum_{k=1}^{5}(k+2)=\sum_{j=0}^{4}(j+3)\). c) The original sum is \(3+4+5+6+7=25\). The incorrect sum is \(2+3+4+5+6=20\), confirming that it represents different terms.

a) The bounds were shifted but the summand was not transformed consistently. b) \(\sum_{j=0}^{4}(j+3)\). c) Original: \(25\); incorrect version: \(20\).

- The two sums cover the same number of terms but use shifted index ranges. - Choose a substitution that turns the lower bound \(2\) into \(1\) and the upper bound \(n+1\) into \(n\). - Combine the sums only after their index variables and bounds have been aligned.

1. In the second sum, let \(j=k+1\). As \(j\) runs from \(2\) to \(n+1\), \(k\) runs from \(1\) to \(n\). 2. Thus \(\sum_{j=2}^{n+1}(j^2-1)=\sum_{k=1}^{n}((k+1)^2-1)\). 3. The original expression becomes \(\sum_{k=1}^{n}\left((k+1)^2-((k+1)^2-1)\right)=\sum_{k=1}^{n}1\). 4. Therefore the expression simplifies to \(n\).

\(n\).

- Reindex the first sum so that its squared terms are written with the same index pattern as the second sum. - Compare the two resulting ranges and identify which middle terms occur in both. - Keep the unmatched terms at the two ends before simplifying algebraically.

1. In the first sum, let \(j=k+2\). Then \(\sum_{k=1}^{n}(k+2)^2=\sum_{j=3}^{n+2}j^2\). 2. Therefore \(T_n=\sum_{j=3}^{n+2}j^2-\sum_{k=1}^{n}k^2\). The terms \(3^2,4^2,\ldots,n^2\) cancel when they are present; the endpoint expression that remains is \(T_n=(n+1)^2+(n+2)^2-1^2-2^2\). This endpoint identity also holds for \(n=1\) and \(n=2\), where the overlap is shorter. 3. Simplifying gives \(T_n=2n^2+6n\).

\(T_n=2n^2+6n\).