{"id":90635,"date":"2025-06-01T10:32:38","date_gmt":"2025-06-01T10:32:38","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=90635"},"modified":"2025-06-01T10:32:38","modified_gmt":"2025-06-01T10:32:38","slug":"suppose-the-nth-term-of-a-series-is-1fracn2fracn22-if","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/suppose-the-nth-term-of-a-series-is-1fracn2fracn22-if\/","title":{"rendered":"Suppose the nth term of a series is $1+\\frac{n}{2}+\\frac{n^2}{2}$. If"},"content":{"rendered":"

Suppose the nth term of a series is $1+\\frac{n}{2}+\\frac{n^2}{2}$. If there are 20 terms in the series, then the sum of the series is equal to<\/p>\n

[amp_mcq option1=”1360″ option2=”1450″ option3=”1500″ option4=”1560″ correct=”option4″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CAPF – 2021<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct answer is D) 1560.
\n<\/section>\n
\nThe nth term of the series is $a_n = 1+\\frac{n}{2}+\\frac{n^2}{2}$. We need to find the sum of the first 20 terms, $S_{20} = \\sum_{n=1}^{20} a_n$.
\n$S_{20} = \\sum_{n=1}^{20} (1+\\frac{n}{2}+\\frac{n^2}{2}) = \\sum_{n=1}^{20} 1 + \\frac{1}{2}\\sum_{n=1}^{20} n + \\frac{1}{2}\\sum_{n=1}^{20} n^2$
\nUsing standard summation formulas:
\n$\\sum_{n=1}^{N} c = Nc$
\n$\\sum_{n=1}^{N} n = \\frac{N(N+1)}{2}$
\n$\\sum_{n=1}^{N} n^2 = \\frac{N(N+1)(2N+1)}{6}$
\nHere, N = 20.
\n$\\sum_{n=1}^{20} 1 = 20 \\times 1 = 20$
\n$\\sum_{n=1}^{20} n = \\frac{20(20+1)}{2} = \\frac{20 \\times 21}{2} = 10 \\times 21 = 210$
\n$\\sum_{n=1}^{20} n^2 = \\frac{20(20+1)(2 \\times 20 + 1)}{6} = \\frac{20 \\times 21 \\times 41}{6} = \\frac{10 \\times 7 \\times 41}{1} = 2870$
\nNow substitute these values back into the sum expression:
\n$S_{20} = 20 + \\frac{1}{2}(210) + \\frac{1}{2}(2870)$
\n$S_{20} = 20 + 105 + 1435$
\n$S_{20} = 125 + 1435 = 1560$.
\n<\/section>\n
\nThe problem requires separating the given nth term into terms based on powers of n and applying standard summation formulas for constants, linear terms, and quadratic terms. Careful calculation is needed for each part and then summing them up.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Suppose the nth term of a series is $1+\\frac{n}{2}+\\frac{n^2}{2}$. If there are 20 terms in the series, then the sum of the series is equal to [amp_mcq option1=”1360″ option2=”1450″ option3=”1500″ option4=”1560″ correct=”option4″] This question was previously asked in UPSC CAPF – 2021 Download PDFAttempt Online The correct answer is D) 1560. The nth term of … <\/p>\n

Detailed SolutionSuppose the nth term of a series is $1+\\frac{n}{2}+\\frac{n^2}{2}$. If<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1085],"tags":[1110,1102],"class_list":["post-90635","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1110","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"\nSuppose the nth term of a series is $1+\\frac{n}{2}+\\frac{n^2}{2}$. If<\/title>\n<meta name=\"description\" content=\"The correct answer is D) 1560. The nth term of the series is $a_n = 1+frac{n}{2}+frac{n^2}{2}$. We need to find the sum of the first 20 terms, $S_{20} = sum_{n=1}^{20} a_n$. $S_{20} = sum_{n=1}^{20} (1+frac{n}{2}+frac{n^2}{2}) = sum_{n=1}^{20} 1 + frac{1}{2}sum_{n=1}^{20} n + frac{1}{2}sum_{n=1}^{20} n^2$ Using standard summation formulas: $sum_{n=1}^{N} c = Nc$ $sum_{n=1}^{N} n = frac{N(N+1)}{2}$ $sum_{n=1}^{N} n^2 = frac{N(N+1)(2N+1)}{6}$ Here, N = 20. $sum_{n=1}^{20} 1 = 20 times 1 = 20$ $sum_{n=1}^{20} n = frac{20(20+1)}{2} = frac{20 times 21}{2} = 10 times 21 = 210$ $sum_{n=1}^{20} n^2 = frac{20(20+1)(2 times 20 + 1)}{6} = frac{20 times 21 times 41}{6} = frac{10 times 7 times 41}{1} = 2870$ Now substitute these values back into the sum expression: $S_{20} = 20 + frac{1}{2}(210) + frac{1}{2}(2870)$ $S_{20} = 20 + 105 + 1435$ $S_{20} = 125 + 1435 = 1560$.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/suppose-the-nth-term-of-a-series-is-1fracn2fracn22-if\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Suppose the nth term of a series is $1+\\frac{n}{2}+\\frac{n^2}{2}$. If\" \/>\n<meta property=\"og:description\" content=\"The correct answer is D) 1560. The nth term of the series is $a_n = 1+frac{n}{2}+frac{n^2}{2}$. We need to find the sum of the first 20 terms, $S_{20} = sum_{n=1}^{20} a_n$. $S_{20} = sum_{n=1}^{20} (1+frac{n}{2}+frac{n^2}{2}) = sum_{n=1}^{20} 1 + frac{1}{2}sum_{n=1}^{20} n + frac{1}{2}sum_{n=1}^{20} n^2$ Using standard summation formulas: $sum_{n=1}^{N} c = Nc$ $sum_{n=1}^{N} n = frac{N(N+1)}{2}$ $sum_{n=1}^{N} n^2 = frac{N(N+1)(2N+1)}{6}$ Here, N = 20. $sum_{n=1}^{20} 1 = 20 times 1 = 20$ $sum_{n=1}^{20} n = frac{20(20+1)}{2} = frac{20 times 21}{2} = 10 times 21 = 210$ $sum_{n=1}^{20} n^2 = frac{20(20+1)(2 times 20 + 1)}{6} = frac{20 times 21 times 41}{6} = frac{10 times 7 times 41}{1} = 2870$ Now substitute these values back into the sum expression: $S_{20} = 20 + frac{1}{2}(210) + frac{1}{2}(2870)$ $S_{20} = 20 + 105 + 1435$ $S_{20} = 125 + 1435 = 1560$.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/suppose-the-nth-term-of-a-series-is-1fracn2fracn22-if\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T10:32:38+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"Suppose the nth term of a series is $1+\\frac{n}{2}+\\frac{n^2}{2}$. If","description":"The correct answer is D) 1560. The nth term of the series is $a_n = 1+frac{n}{2}+frac{n^2}{2}$. We need to find the sum of the first 20 terms, $S_{20} = sum_{n=1}^{20} a_n$. $S_{20} = sum_{n=1}^{20} (1+frac{n}{2}+frac{n^2}{2}) = sum_{n=1}^{20} 1 + frac{1}{2}sum_{n=1}^{20} n + frac{1}{2}sum_{n=1}^{20} n^2$ Using standard summation formulas: $sum_{n=1}^{N} c = Nc$ $sum_{n=1}^{N} n = frac{N(N+1)}{2}$ $sum_{n=1}^{N} n^2 = frac{N(N+1)(2N+1)}{6}$ Here, N = 20. $sum_{n=1}^{20} 1 = 20 times 1 = 20$ $sum_{n=1}^{20} n = frac{20(20+1)}{2} = frac{20 times 21}{2} = 10 times 21 = 210$ $sum_{n=1}^{20} n^2 = frac{20(20+1)(2 times 20 + 1)}{6} = frac{20 times 21 times 41}{6} = frac{10 times 7 times 41}{1} = 2870$ Now substitute these values back into the sum expression: $S_{20} = 20 + frac{1}{2}(210) + frac{1}{2}(2870)$ $S_{20} = 20 + 105 + 1435$ $S_{20} = 125 + 1435 = 1560$.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/suppose-the-nth-term-of-a-series-is-1fracn2fracn22-if\/","og_locale":"en_US","og_type":"article","og_title":"Suppose the nth term of a series is $1+\\frac{n}{2}+\\frac{n^2}{2}$. If","og_description":"The correct answer is D) 1560. The nth term of the series is $a_n = 1+frac{n}{2}+frac{n^2}{2}$. We need to find the sum of the first 20 terms, $S_{20} = sum_{n=1}^{20} a_n$. $S_{20} = sum_{n=1}^{20} (1+frac{n}{2}+frac{n^2}{2}) = sum_{n=1}^{20} 1 + frac{1}{2}sum_{n=1}^{20} n + frac{1}{2}sum_{n=1}^{20} n^2$ Using standard summation formulas: $sum_{n=1}^{N} c = Nc$ $sum_{n=1}^{N} n = frac{N(N+1)}{2}$ $sum_{n=1}^{N} n^2 = frac{N(N+1)(2N+1)}{6}$ Here, N = 20. $sum_{n=1}^{20} 1 = 20 times 1 = 20$ $sum_{n=1}^{20} n = frac{20(20+1)}{2} = frac{20 times 21}{2} = 10 times 21 = 210$ $sum_{n=1}^{20} n^2 = frac{20(20+1)(2 times 20 + 1)}{6} = frac{20 times 21 times 41}{6} = frac{10 times 7 times 41}{1} = 2870$ Now substitute these values back into the sum expression: $S_{20} = 20 + frac{1}{2}(210) + frac{1}{2}(2870)$ $S_{20} = 20 + 105 + 1435$ $S_{20} = 125 + 1435 = 1560$.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/suppose-the-nth-term-of-a-series-is-1fracn2fracn22-if\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T10:32:38+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/suppose-the-nth-term-of-a-series-is-1fracn2fracn22-if\/","url":"https:\/\/exam.pscnotes.com\/mcq\/suppose-the-nth-term-of-a-series-is-1fracn2fracn22-if\/","name":"Suppose the nth term of a series is $1+\\frac{n}{2}+\\frac{n^2}{2}$. If","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T10:32:38+00:00","dateModified":"2025-06-01T10:32:38+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct answer is D) 1560. The nth term of the series is $a_n = 1+\\frac{n}{2}+\\frac{n^2}{2}$. We need to find the sum of the first 20 terms, $S_{20} = \\sum_{n=1}^{20} a_n$. $S_{20} = \\sum_{n=1}^{20} (1+\\frac{n}{2}+\\frac{n^2}{2}) = \\sum_{n=1}^{20} 1 + \\frac{1}{2}\\sum_{n=1}^{20} n + \\frac{1}{2}\\sum_{n=1}^{20} n^2$ Using standard summation formulas: $\\sum_{n=1}^{N} c = Nc$ $\\sum_{n=1}^{N} n = \\frac{N(N+1)}{2}$ $\\sum_{n=1}^{N} n^2 = \\frac{N(N+1)(2N+1)}{6}$ Here, N = 20. $\\sum_{n=1}^{20} 1 = 20 \\times 1 = 20$ $\\sum_{n=1}^{20} n = \\frac{20(20+1)}{2} = \\frac{20 \\times 21}{2} = 10 \\times 21 = 210$ $\\sum_{n=1}^{20} n^2 = \\frac{20(20+1)(2 \\times 20 + 1)}{6} = \\frac{20 \\times 21 \\times 41}{6} = \\frac{10 \\times 7 \\times 41}{1} = 2870$ Now substitute these values back into the sum expression: $S_{20} = 20 + \\frac{1}{2}(210) + \\frac{1}{2}(2870)$ $S_{20} = 20 + 105 + 1435$ $S_{20} = 125 + 1435 = 1560$.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/suppose-the-nth-term-of-a-series-is-1fracn2fracn22-if\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/suppose-the-nth-term-of-a-series-is-1fracn2fracn22-if\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/suppose-the-nth-term-of-a-series-is-1fracn2fracn22-if\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CAPF","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-capf\/"},{"@type":"ListItem","position":3,"name":"Suppose the nth term of a series is $1+\\frac{n}{2}+\\frac{n^2}{2}$. If"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/90635","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=90635"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/90635\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=90635"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=90635"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=90635"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}