Question:
\((S1) : lim_{n→∞}\frac{1}{ n ^2} ( 2 + 4 + 6 + . . . . + 2n) = 1\)
\((S2): lim_n→∞\frac{1}{n^{16}}( 1^15 +2^15 +3^15 + . . . . + n^15 ) = \frac{1}{16} \)
\((S1) : lim_{n→∞}\frac{1}{ n ^2} ( 2 + 4 + 6 + . . . . + 2n) = 1\)
\((S2): lim_n→∞\frac{1}{n^{16}}( 1^15 +2^15 +3^15 + . . . . + n^15 ) = \frac{1}{16} \)
\((S2): lim_n→∞\frac{1}{n^{16}}( 1^15 +2^15 +3^15 + . . . . + n^15 ) = \frac{1}{16} \)
Updated On: Sep 14, 2024
- Only (S1) is true
- Only (S2) is true
- Both (S1) and (S2) are true
- Both (S1) and (S2) are false
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The Correct Option is C
Solution and Explanation
The correct option is(C): Only (S2) is true
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