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1, 4, 9, 16, ………………. sum to ‘n’ terms =A) \(\cfrac{n^2(n+1)^2}4\)B) \(\cfrac{n(n+1)(2n+1)}6\)C) \(\cfrac{(n+1)(2n+1)}3\)D) \(\cfrac{n(n+1)}2\) |
Answer» Correct option is B) \(\cfrac{n(n+1)(2n+1)}6\) |
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