Q.
The identity 13+23+33+…+n3 is equal to
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a
n(n−1)22
b
n(n+1)2
c
{n(n+1)}22
d
n(n+1)22
answer is D.
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Detailed Solution
Let the given statement be P(n).P(n):13+23+33+…+n3=n(n+1)22Step I : For n=1, P(1):1(1+1)22=1×222=12=1=13 which is trueStep ll: Let it is true for n = k, 13+23+33+…+k3=k(k+1)22------iStep lll: For n=k+1, 13+23+33+43+…+k3+(k+1)3 =k(k+1)22+(k+1)3 [using Eq. (i)] =k2(k+1)24+(k+1)31=k2(k+1)2+4(k+1)34 On taking (k + 1)2 common in numerator Part, =(k+1)2k2+4(k+1)4=(k+1)2k2+4k+44=(k+1)2(k+2)24=(k+1)2[(k+1)+1]24=(k+1){(k+1)+1}22Therefore, P(k + 1) is true when P(k) is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers n.
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