Free math worksheets

- Read the symbol directly below and above \(\sum\). - Separate the changing index from the expression being added. - Count the integer index values inclusively from the lower bound to the upper bound.

The index variable is \(k\). The lower bound is \(2\), the upper bound is \(6\), and the summand is \(3k+1\). The index values are \(2,3,4,5,6\), so there are \(5\) terms.

Index: \(k\); lower bound: \(2\); upper bound: \(6\); summand: \(3k+1\); number of terms: \(5\)

- Start with the lower-bound value of the index and increase it by \(1\) each time. - Substitute each allowed index into the summand in order.

Substitute \(j=1,2,3,4\) into \(2j-1\). The expanded sum is \(1+3+5+7\).

\(1+3+5+7\)

- Expand the sigma notation from the lower bound through the upper bound. - Evaluate each power only after listing all included terms.

The sum is \(2^0+2^1+2^2+2^3+2^4=1+2+4+8+16=31\).

\(31\)

- Identify a formula for the \(k\)th term of the arithmetic pattern. - Determine how many terms are included before choosing the upper bound. - Check your sigma form by substituting the first and last index values.

The terms form an arithmetic sequence with first term \(4\) and common difference \(3\). Since \(31=4+3(10-1)\), there are \(10\) terms. A suitable form is \(\sum_{k=1}^{10}[4+3(k-1)]\).

\(\sum_{k=1}^{10}[4+3(k-1)]\)

- The summand does not affect how many index values are used. - Count the integers from the lower bound through the upper bound inclusively.

The integer index values run from \(-3\) through \(7\), inclusive. The number of terms is \(7-(-3)+1=11\).

\(11\) terms

- Identify the factor that does not depend on the index. - Move that constant outside the sigma symbol before evaluating the simpler sum.

Factor out the constant: \(\sum_{k=1}^{6}5k=5\sum_{k=1}^{6}k\). Since \(1+2+3+4+5+6=21\), the value is \(5\cdot21=105\).

\(105\)

- Separate the two terms of the summand before evaluating. - Keep the same index bounds on each resulting sigma expression. - Apply any constant factor only to the part where it belongs.

Split the summand: \(\sum_{k=1}^{4}(k^2+2k)=\sum_{k=1}^{4}k^2+2\sum_{k=1}^{4}k\). The first sum is \(1+4+9+16=30\), and the second is \(2(1+2+3+4)=20\). Therefore, the total is \(50\).

\(\sum_{k=1}^{4}k^2+2\sum_{k=1}^{4}k=50\)

- Identify the first term and the multiplicative factor between consecutive terms. - Use an exponent that is \(0\) when the index is at its first value. - Make the upper bound match the number of displayed terms.

The terms form a geometric sequence with first term \(3\) and common ratio \(2\). With five terms, the sum is \(\sum_{k=1}^{5}3\cdot2^{k-1}\).

\(\sum_{k=1}^{5}3\cdot2^{k-1}\)

- List every integer index from the lower bound through the upper bound. - Substitute each index into the summand before adding. - Check that the number of expanded terms matches the number of allowed index values.

Substituting \(k=3,4,5,6,7\) gives \(11+15+19+23+27=95\).

\(95\)

- Compare the starting values of the two indices. - Choose an index shift that makes the lower bounds correspond. - Transform both the bounds and the summand; changing only one is not enough.

Let \(j=k-1\), so \(k=j+1\). When \(k=1\), \(j=0\); when \(k=6\), \(j=5\). Also, \(3k-2=3(j+1)-2=3j+1\). Therefore, \(\sum_{k=1}^{6}(3k-2)=\sum_{j=0}^{5}(3j+1)\).

They are equal under the re-indexing \(j=k-1\).

- Test the proposed summand at its first index value. - Determine how many arithmetic-sequence terms run from the first displayed value through the last. - A correct sigma expression must reproduce both endpoints and the correct term count.

The proposed summand gives \(8\) when \(k=1\), so it starts at the second term instead of \(5\). It also uses only \(9\) index values, while the arithmetic sequence from \(5\) to \(32\) has \(10\) terms. A correct form is \(\sum_{k=1}^{10}[5+3(k-1)]\).

The proposal starts one term late and includes only \(9\) terms. One correct expression is \(\sum_{k=1}^{10}[5+3(k-1)]\).

- Express the number of integer index values from the lower bound through \(14\). - Remember that inclusive counting adds one after subtracting the bounds. - Solve the resulting equation for the unknown lower bound.

An inclusive sum from \(p\) through \(14\) has \(14-p+1\) terms. Set \(14-p+1=9\). Then \(15-p=9\), so \(p=6\).

\(p=6\)

- Interpret the summand values as a finite arithmetic sequence. - Identify its first and last terms in terms of \(n\). - Keep the final result symbolic rather than substituting a particular value of \(n\).

The summand produces the arithmetic sequence \(1,5,9,\ldots,4n-3\). Its sum is \(\frac n2[1+(4n-3)]\) \(=\frac n2(4n-2)=n(2n-1)\).

\(n(2n-1)\)

- Try shifting the index of \(A\) so its lower bound becomes \(0\). - When re-indexing, transform the summand and both bounds together. - Compare the first generated term of each expression before deciding whether the sums can match.

Re-index \(A\) by setting \(j=k-2\). Then \(k=j+2\), so \(4k-5=4(j+2)-5=4j+3\). The lower bound \(k=2\) becomes \(j=0\), and the upper bound \(k=n+1\) becomes \(j=n-1\). Therefore, \(A=B\). Expression \(C\) uses the same form \(4m+3\) but starts at \(m=1\), so its first term is \(7\) instead of \(3\). Thus, \(C\) is not equivalent to \(A\) and \(B\).

\(A=B\) for every positive integer \(n\). Expression \(C\) is different because its indexing starts one step later.