Sum of first n natural numbers

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{135}=\frac{135\times136}{2}=9180\) and \(S_{90}=\frac{90\times91}{2}=4095\). Therefore, the difference is \(9180-4095=5085\). Equivalently, it is the sum of the numbers from 91 to 135. Exam tip: write \(S_{135}-S_{90}\) first, then apply the sum formula.

The direct answer is option D, \(S_{120}\). By definition, \(S_{115}=1+2+\cdots+115\). Adding the next consecutive terms \(116,117,118,119,120\) extends the sum without skipping any number: \(S_{115}+116+117+118+119+120=1+2+\cdots+120=S_{120}\). Option D is correct. Option A, \(S_{117}\), would include only 116 and 117 after \(S_{115}\), leaving 118–120 out. Option B, \(S_{118}\), leaves 119 and 120 out. Option C, \(S_{119}\), leaves 120 out. The final added term determines the new upper index because all intermediate terms are also added. Memory cue: start at \(S_k\), add every next term through \(m\), and obtain \(S_m\).

(S_{125}=\frac{125\times126}{2}=7875), so (n=125). For large values, using (2S_n) makes identification easier.

(S_{29}=435), (S_{58}=1711), and (S_{87}=3828), so the total is (5974). Add the three different indices carefully.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Hence, \(\frac{n(n+1)}{2}=10878\), so \(n(n+1)=21756\). Since \(147\times148=21756\), \(n=147\). Therefore, \(n-25=147-25=122\). Thus, option C is correct. A nearby choice such as 120 would give \(n=145\), whose sum is not 10878. Exam tip: after using \(n(n+1)/2\), look for two consecutive integers whose product equals twice the given sum.

Answer: D, 71. The sum Sₙ₊₂ contains all numbers from 1 through n + 2, while Sₙ₋₂ contains all numbers from 1 through n − 2. Subtracting cancels the common terms 1 through n − 2. The four terms left are n − 1, n, n + 1, and n + 2. Their sum is (n − 1) + n + (n + 1) + (n + 2) = 4n + 2. Equating this with 286 gives 4n + 2 = 286, so 4n = 284 and n = 71. Thus option D is correct. A direct check is helpful: if n = 71, the remaining terms are 70, 71, 72, and 73, whose sum is 70 + 71 + 72 + 73 = 286. Option A gives 68, option B gives 69, and option C gives 70; each produces a four-term sum smaller than 286. The key counting point is that the index gap is four, so four terms remain.

Answer: C, 7626. Use Sᵧ = y(y + 1)/2. We need y(y + 1)/2 = 7381. Since 121 × 122/2 = 121 × 61 = 7381, y = 121. Moving from Sᵧ to Sᵧ₊₂ does not add y and y + 1; y is already included in Sᵧ. It adds the next two terms, y + 1 and y + 2, which are 122 and 123. Therefore Sᵧ₊₂ = Sᵧ + 122 + 123 = 7381 + 245 = 7626. Hence option C is correct. Option A, 7586, comes from an incorrect addition or omission. Option B may result from adding only part of the required increase, and option D is too large because it adds an incorrect amount. The important indexing rule is Sᵣ₊₂ − Sᵣ = (r + 1) + (r + 2), because the original last term r is already present.

Answer: D, 5184. Apply Sₙ = n(n + 1)/2 to each partial sum. First, S₁₄₄ = 144 × 145/2 = 72 × 145 = 10440. Next, S₇₂ = 72 × 73/2 = 36 × 73 = 2628, so 2S₇₂ = 2 × 2628 = 5256. Therefore S₁₄₄ − 2S₇₂ = 10440 − 5256 = 5184. Thus option D is correct. Option A, B, and C do not result from the correct substitution and operation. A likely error is to forget the coefficient 2 and subtract only S₇₂, which would give 7812, not one of the listed answers; another error is to treat 2S₇₂ as S₁₄₄ or as S₁₄₄₋₂. The coefficient 2 multiplies the complete value of S₇₂, not its index. Calculate each sum first, then apply the subtraction.

Answer: D, 38.5. Let m = 2k + 3. Then the given statement becomes Sₘ = 3240. Using Sₘ = m(m + 1)/2, we get m(m + 1)/2 = 3240, so m(m + 1) = 6480. The consecutive integers 80 and 81 have product 80 × 81 = 6480, so m = 80. Now substitute back: 2k + 3 = 80, hence 2k = 77 and k = 77/2 = 38.5. Therefore option D is correct. It is not necessary for k itself to be a whole number; the important index 2k + 3 equals 80, a valid natural-number index. Option A gives index 77, option B gives index 78, and option C gives index 79; none has sum 3240. A common confusion is to assume every variable in a sum problem must be an integer. Here the index is integral, while k may be fractional.

Treat S_n and S_(n−1) as two unknown quantities. Adding the two given equations eliminates S_(n−1): (S_n+S_(n−1))+(S_n−S_(n−1))=4212+65, so 2S_n=4277. Dividing by 2 gives S_n=2138.5, which is option A. Subtracting the equations would give 2S_(n−1)=4147, confirming the algebra. There is a noteworthy consistency issue: if S_n denotes the sum of the first n natural numbers, it should be an integer, so the supplied data cannot correspond to an actual natural-number index n. Nevertheless, the simultaneous equations uniquely determine the listed value.

First determine n from the triangular-number equation n(n+1)/2=7140. Since 119×120/2=7140, n=119. Now interpret the difference carefully: S_(n+5)−S_n contains exactly the five terms after n, namely 120, 121, 122, 123, and 124. Their sum is an arithmetic-series sum: number of terms × average =5×(120+124)/2=5×122=610. Therefore option D is correct. The value 605 in option C is not supported by the calculation. The key indexing point is that n itself is already included in S_n and cancels, while the next five numbers remain. This explains both the answer and the main distractor.