📝 Sigma notation summation explained (24 MCQs)

📖 Explanation: The summand $(2k)^2$ generates the terms $2^2, 4^2, 6^2, \dots$ as $k = 1,2,\dots,n$. Option B incorrectly sums $k^2$ from 1 to $2n$, which would include odd squares. Option C incorrectly factors out the 2 from the square, and Option D omits the square entirely.

📖 Explanation: Using the distributive property of sigma: $\sum (3a_k - 2b_k) = 3 \sum a_k - 2 \sum b_k = 3(45) - 2(-12) = 135 + 24 = 159$. Option A incorrectly treats $-2b_k$ as subtraction of positive 24. Option C incorrectly adds 24 instead of subtracting negative 24. Option D incorrectly treats the sum of the sums as $45 - (-12)$.

📖 Explanation: The correct value is $1^2+2^2+3^2+4^2+5^2 = 55$. The formula $n(n+1)/2 = 15$ gives the sum of the first 5 integers, not squares. A student getting 50 likely misapplied the formula for sum of squares or made an arithmetic error. Option B would give 15, Option C gives 45, Option D gives 225.

📖 Explanation: The lower limit is 2 and the upper limit is 5, so the summand is evaluated at $k=2,3,4,5$. The correct sum is $(4-3)+(9-3)+(16-3)+(25-3) = 1+6+13+22 = 42$. Option A starts from $k=2$ but incorrectly lists 4 as the first square. Option C uses $2^2-3=1$ and $3^2-3=6$, but then sums incorrectly. Option D is an overestimation.

📖 Explanation: The sum $\sum_{k=1}^{n} (2k-1)$ represents the sum of the first $n$ odd positive integers, which equals $n^2$. This can be verified for small $n$: for $n=3$, $1+3+5 = 9 = 3^2$. Option B is the sum of the first $n$ integers. Option C is $n^2-1$, which would be the sum of odd numbers except the first? No, it's incorrect. Option D doubles the square.

📖 Explanation: The index of summation is a dummy variable. Changing $k$ to $j$ or any other letter not used elsewhere does not change the sum. The number of terms from $k=m$ to $n$ is $n-m+1$, not necessarily $n+1$.

📖 Explanation: Using the sum rule: $\sum_{i=1}^{10} (i+1) = \sum_{i=1}^{10} i + \sum_{i=1}^{10} 1 = 55 + 10 = 65$. Option B is a common mistake of adding 1 to the sum instead of adding 1 for each term. Option C is correct. Option D adds 11 because they might think the number of terms is 11.

📖 Explanation: The formula $\frac{n^2(n+1)^2}{4}$ is the closed form for the sum of cubes, $\sum_{k=1}^{n} k^3$. The correct formula for $\sum k^2$ is $\frac{n(n+1)(2n+1)}{6}$. The student has confused the two.

📖 Explanation: $\sum_{k=1}^{5} (2k+1) = \sum_{k=1}^{5} (2k-1+2) = \sum_{k=1}^{5} (2k-1) + \sum_{k=1}^{5} 2 = 25 + 10 = 35$. Option C is a common error where students add 5 (for the number of terms) instead of 2 per term. Option B adds only 2, ignoring the fact that the constant is added to each of the 5 terms.

📖 Explanation: If $j = k-1$, then when $k=1$, $j=0$; when $k=n$, $j=n-1$. The summand becomes $a_{j+1}$. So $\sum_{k=1}^{n} a_k = \sum_{j=0}^{n-1} a_{j+1}$. Option A omits the ' +1 ' in the subscript, incorrectly changing the terms.

📖 Explanation: The number of terms is $10 - 3 + 1 = 8$. Students often subtract 3 from 10 to get 7 (Option A), forgetting to add 1. Option C and D are incorrect.

📖 Explanation: The lower limit is 0, so the terms are $a_0 = 2^0 = 1$, $a_1 = 2$, $a_2 = 4$, $a_3 = 8$, and $a_4 = 16$. Sum is 31. Option B starts with $k=1$. Option C has an extra term $2^5$. Option D includes $k=0$ but incorrectly uses 0 instead of $2^0$.

📖 Explanation: $k(k+1) = k^2 + k$. Sum = $\sum k^2 + \sum k = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)}{6} [(2n+1)+3] = \frac{n(n+1)(2n+4)}{6} = \frac{n(n+1)(n+2)}{3}$. The student's answer is correct. Option B is the formula for $\sum k^2$ only. Option C is the formula for $\sum k(k+1)(k+2)/3$.

📖 Explanation: $\sum_{k=1}^{10} \frac{k(k+1)}{2} = \frac{1}{2} \sum (k^2 + k) = \frac{1}{2} [ \frac{10\cdot 11 \cdot 21}{6} + \frac{10\cdot 11}{2} ] = \frac{1}{2} [385 + 55] = 220$. Option B is incorrect algebra. Option C is the sum of squares, not the correct formula. Option D uses the sum of first 10 integers incorrectly.

📖 Explanation: $k=1$ gives $1/2^1 = 1/2$, $k=2$ gives $1/4$, $k=3$ gives $1/8$, $k=4$ gives $1/16$. Option A incorrectly has denominators 2,4,6,8. Option C is sum of reciprocals of integers. Option D starts at 2.

📖 Explanation: The formula is correct for sum of cubes. For $n=5$, $\frac{5^2 \cdot 6^2}{4} = \frac{25 \cdot 36}{4} = 225$. Option B is $1^3+2^3+3^3+4^3 = 100$. Option C is $1^3+2^3+3^3+4^3+5^3 = 225$.

📖 Explanation: The sum $\sum (a_k + b_k)^2 = \sum (a_k^2 + 2a_k b_k + b_k^2)$. Without knowing $\sum a_k^2$, $\sum b_k^2$, and $\sum a_k b_k$, we cannot determine the value from just the sums of $a_k$ and $b_k$. Option B incorrectly squares the sums. Option C adds the sums. Option D treats the sum of the squared sum as the square of the sum, which is false.

📖 Explanation: $\sum_{k=1}^{n} (2k+1) = 2 \sum k + \sum 1 = 2 \cdot \frac{n(n+1)}{2} + n = n(n+1) + n = n^2 + n + n = n^2 + 2n = n(n+2)$. Option B is $n^2 + 1$, which is incorrect. Option C is $n^2 + n$. Option D is $n^2 + 2n$.

📖 Explanation: $\sum_{k=1}^{n} (k^2 - k) = \sum k^2 - \sum k = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} = \frac{n(n+1)}{6} [(2n+1) - 3] = \frac{n(n+1)}{6} (2n-2) = \frac{n(n+1)(n-1)}{3}$. Option B is sum of squares. Option C is sum of first n-1 integers? No, $n(n-1)/2$ is sum of first n-1 integers.

📖 Explanation: $k=1: \sin(\pi/2)=1$, $k=2: \sin(\pi)=0$, $k=3: \sin(3\pi/2)=-1$, $k=4: \sin(2\pi)=0$. Sum = 0. Option B is just the first term. Option D is sum of 1 and 1? No.

📖 Explanation: The sum $\sum_{k=1}^{n} 1$ has $n$ terms, each equal to 1, so the sum is $n$. Option B would be a single term. Option C is for sum from 0 to n.

📖 Explanation: The summand is $k / 2^k$. For k=1, 1/2; k=2, 2/4; k=3, 3/8; k=4, 4/16; k=5, 5/32. Option B misses the exponent. Option C has numerator 1 for all terms. Option D squares the numerator, which is incorrect.