Find the value of “n” where 348 + 31996 + 33943 + 33n. Options:
A) 1963 B) 1964 C) 1960 D) 1991 E) 1992
3^48 + 3^1996 + 3^3943+3^3n   (3^16 )^3 + 3^1996 + 3^3943 + (3^n)^3   its expansion of (a+b)^3=a^3+3a^2b+3ab^2+b^3   =&gt; 3*(3^16)^2*3^n=3^1996   =&gt; 3^(33+n)=3^1996   =&gt; n = 1963   1963NSWER:1963
<result>A</result>
Find the value of “n” where 348 + 31996 + 33943 + 33n.